RD Sharma Solutions Class 12 Mathematics Chapter 21 RE

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 05:45 PM IST

RD Sharma Solutions for Class 12 are made to help students perform well and learn various concepts. RD Sharma books have a good reputation for being informative and comprehensive. Class 12 Maths is a challenging subject that requires a lot of practice. RD Sharma Class 12th Exercise RE material helps students practice better as it contains solved examples that have step-by-step answers. This is an excellent alternative for preparation as it follows the CBSE syllabus and is prepared by subject experts. RD Sharma Class 12 Chapter 21 Exercise RE also helps students with Revision as the solutions are easy to understand and are available in one place.

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RD Sharma Class 12 Solutions Chapter21 RE Differential Equation- Other Exercise

Differential Equations Excercise:RE

Differential Equations Exercise Revision Exercise Question 1(i)

Answer:
Order $-2$ , degree $-1$
Hint:
Check the highest order of derivate to find the order and check the power of derivative to find degree.
Given:
$\left ( \frac{ds}{dt} \right )^{4}+3s\left ( \frac{d^{2}s}{dt^{2}} \right )=0$
Solution:
$\left ( \frac{ds}{dt} \right )^{4}+3s\left ( \frac{d^{2}s}{dt^{2}} \right )=0$
$\therefore \left ( s' \right )^{4}+3s\left ( s^{n} \right )=0$

Highest order derivative $=2$

$\therefore$ Order $=2$

Degree= Power of $s"$

Degree $=1$

Differential Equations Exercise Revision Exercise Question 1(ii)

Answer:
Order $-3$ ,Degree $-1$
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
${y}''{}'+{2y}''+{y}'=0$
Solution:
${y}''{}'+{2y}''+{y}'=0$
Highest order derivative $=3$
$\therefore$ Order $=3$
Degree $=$ Power of ${{y}'}''$
Degree $=1$

Differential Equations Exercise Revision Exercise Question 1(iii)

Answer:
Degree $-2$ ,Order $-3$
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
$\left ( {{y}'}'' \right )^{2}+\left ( y{}'' \right )^{3}+\left ( {y}' \right )^{4}+y^{5}=0$
Solution:
$\left ( {{y}'}'' \right )^{2}+\left ( y{}'' \right )^{3}+\left ( {y}' \right )^{4}+y^{5}=0$
Highest order derivative $=3$
$\therefore$ Order $=3$
Degree $=$ power of $\left ( {{y}'}'' \right )^{2}$
$\therefore$ Degree $=2$

Differential Equations Exercise Revision Exercise Question 1(iv)

Answer:
Degree $-1$ ,Order $-3$
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
${{y}'}''+2{y}''+{y}'=0$
Solution:
${{y}'}''+2{y}''+{y}'=0$
Highest order derivative $=3$
$\therefore$ Order $=3$
Degree $=$ power of $\left ( {{y}'}'' \right )^{1}$
$\therefore$ Degree $=1$

Differential Equations Exercise Revision Exercise Question 1(v)

Answer:
Degree $-1$ ,Order $-2$
Hint:

Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
${y}''+\left ( {y}' \right )^{2}+2y=0$
Solution:
${y}''+\left ( {y}' \right )^{2}+2y=0$
Highest order derivative $=2$
$\therefore$ Order $=2$
Degree $=$ power of $\left ( {y}' \right )^{1}$
$\therefore$ Degree $=1$

Differential Equations Exercise Revision Exercise Question 1(vii)

Answer:
Degree is not defined, order $-3$
Hint:
Check the highest order of derivate to find the order and check the power of highest derivative to find degree.
Given:
${{y}'}''+y^{2}+e^{{y}'}=0$
Solution:
${{y}'}''+y^{2}+e^{{y}'}=0$
Highest order derivative $=3$
$\therefore$ Order $=3$
As the equation cannot be expressed as a polynomial of derivative
$\therefore$ Degree is not defined

Differential Equations Exercise Revision Exercise Question 2

Answer:
Verified
Hint:
Find double derivatives of given equation and put values to verify
Given:
$y=e^{-3x}$
Prove that $y=e^{-3x}$ is the solution of $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0$
Solution:
$y=e^{-3x}$
$\begin{aligned} &\frac{d y}{d x}=-3 e^{-3 x} \\ &\frac{d^{2} y}{d x^{2}}=(-3)(-3) e^{-3 x} \\ &\frac{d^{2} y}{d x^{2}}=9 e^{-3 x} \end{aligned}$
Put it in equation
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0 \\ &9 e^{-3 x}-3 e^{-3 x}-6 e^{-3 x}=0 \\ &\Rightarrow 0 \end{aligned}$
Hence proved.

Differential Equations Exercise Revision Exercise Question 3(i)

Answer:
Verified
Hint:
Find the first and second derivative of given function and put values in differential equation to verify
Given:
$y=e^{x}+1$
${y}''-{y}'=0$
Solution:
$\begin{aligned} &y=e^{x}+1 \\ &\frac{d y}{d x}=e^{x} \\ &\frac{d^{2} y}{d x^{2}}=e^{x} \end{aligned}$
Put in given differential equation,
$\begin{aligned} &y^{\prime \prime}-y^{\prime}=0 \\ &\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0 \\ &e^{x}-e^{x}=0 \\ &\Rightarrow 0 \end{aligned}$
Hence verified

Differential Equations Exercise Revision Exercise Question 3(ii)

Answer:

Verified

Hint:
Find the first and second derivative of given function and put in differential equation to verify
Given:
$y=x^{2}+2x+c$
${y}'-2x-2=0$
Solution:
$\begin{aligned} y &=x^{2}+2 x+c \\ y^{\prime} &=\frac{d y}{d x}=2 x+2 \\ y^{\prime \prime} &=\frac{d^{2} y}{d x^{2}}=2 \end{aligned}$
Put in differential equation,
${y}'-2x-2=0$
$2x+2-2x-2=0$
$\Rightarrow 0$
Hence verified

Differential Equations Exercise Revision Exercise Question 3(iii)

Answer:
Verified
Hint:
Find the first and second derivative of given function and put in differential equation to verify
Given:
$y=\cos x+c$
${y}'+\sin x=0$
Solution:
$y=\cos x+c$
${y}'=\frac{dy}{dx}=-\sin x$
Put in differential equation,
${y}'+\sin x=0$
$-\sin x+\sin x=0$
$\Rightarrow 0$
Hence verified

Differential Equations Exercise Revision Exercise Question 3(iv)

Answer:
Verified
Hint:
Find the first derivative of given function and put value in differential equation
Given:
$y=\sqrt{1+x^{2}}$
${y}'=\frac{xy}{1+x^{2}}$
Solution:
$\begin{aligned} &y^{\prime}=\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{1+x^{2}}} \cdot 2 x \ldots \text { using chain rule } \\ &y^{\prime}=\frac{x}{\sqrt{1+x^{2}}} \end{aligned}$
Put in differential equation,
${y}'=\frac{xy}{1+x^{2}}$
LHS $=$ $\frac{x}{\sqrt{1+x^{2}}}$
Multiply and divide by $\sqrt{1+x^{2}}$
$\begin{aligned} &=\frac{x}{\sqrt{1+x^{2}}} \times \frac{\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}} \\ &=\frac{x \sqrt{1+x^{2}}}{\left(1+x^{2}\right)} \end{aligned}$
Where, $y=\sqrt{1+x^{2}}$
Therefore, LHS $= \frac{xy}{1+x^{2}}$ RHS..Hence verified.

Differential Equations Exercise Revision Exercise Question 3(v)

Answer:
Verified
Hint:
You must firstly solve the first derivative and put value in differential equation
Given:
$y=x\sin x$
${xy}'=y+x\sqrt{x^{2}-y^{2}}$
Solution:
${y}'=\frac{dy}{dx}=\sin x+x\cos x.....using \; \; product\: \: rule$
Put in differential equation ${xy}'=y+x\sqrt{x^{2}-y^{2}}$
$LHS=x y^{\prime}=x(\sin x+x \cos x)$
$=x \sin x+x^{2} \cos x$
$\begin{aligned} &\text { RHS }=y+x \sqrt{x^{2}-y^{2}}\\ &=x \sin x+x \sqrt{x^{2}-(x \sin x)^{2}}\\ &\begin{aligned} &=x \sin x+x \sqrt{x^{2}\left(1-\sin ^{2} x\right)} \\ &=x \sin x+x^{2} \sqrt{\cos ^{2} x} \end{aligned} \quad\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \cos ^{2} x=1-\sin ^{2} x \end{array}\right]\\ &=x \sin x+x^{2} \cos x \end{aligned}$
LHS $=$ RHS
Hence verified

Differential Equations Exercise Revision Exercise Question 3(vi)

Answer:
Verified
Hint:
Find first derivative and put value in differential equation to verify
Given:
$y=\sqrt{a^{2}-x^{2}}$
$x+y\frac{dy}{dx}=0$
Solution:
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{2} \frac{1}{\sqrt{a^{2}-x^{2}}}(-2 x) \ldots \text { using chain rule } \\ &\frac{d y}{d x}=\frac{-x}{\sqrt{a^{2}-x^{2}}} \end{aligned}$
Put in differential equation $x+y\frac{dy}{dx}$
$\begin{aligned} &=x+\sqrt{a^{2}-x^{2}}\left(\frac{-x}{\sqrt{a^{2}-x^{2}}}\right) \\ &=x-x \\ &=0 \end{aligned}$
$= RHS$
Hence verified

Differential Equations Exercise Revision Exercise Question 4

Answer:
$x\frac{dy}{dx}-y=0$
Hint:
You must have the knowledge about the curves and make differential equation.
Given:
$y=mx$
Where, $m$ is an arbitrary constant
Solution:
$y=mx$
$\frac{dy}{dx}=m$
Substitute $m=\frac{dy}{dx}$ in $y=mx$
$\therefore$ $y=\frac{d y}{d x}(x) \\$
$\frac{d y}{d x}=\frac{y}{x}$
$\frac{d y}{d x}-\frac{y}{x}=0 \\$
$x \frac{d y}{d x}-y=0$ which is the required equation

Differential Equations Exercise Revision Exercise Question 5

Answer:
$\frac{d^{2}y}{dx^{2}}+y=0$
Hint:
The number of constants is equal to the number of time we differentiate
Given:
$y=a\: \sin \left ( x+b \right )$ , where $a,b$ are constants
Solution:
$y=a\: \sin \left ( x+b \right )$
Here, there are two constants. So, we differentiate twice
$\frac{dy}{dx}=a\cos \left ( x+b \right )$
Again,
$\frac{d^{2}y}{dx^{2}}=-a\sin \left ( x+b \right )$
$\frac{d^{2}y}{dx^{2}}=-y$
$\therefore$ $\frac{d^{2}y}{dx^{2}}+y=0$ which is the required differential equation.

Differential Equations Exercise Revision Exercise Question 6

Answer:
$y^{2}-2xy{y}'=0$
Hint:
You must know the equation of parabola
Given:
Family of parabola having vertex at origin and axis along positive direction of x-axis
Solution:
Since parabola has axis along positive x-axis , its equation is
$y^{2}=4ax$ … (i)
Differentiate,
$2y\frac{dy}{dx}=4a$
Putting value in (i)
$y^{2}=2y\frac{dy}{dx}\left ( x \right )$
$y^{2}-2xy\frac{dy}{dx}=0$ which is the required differential equation

Differential Equations Exercise Revision Exercise Question 7

Answer:
$\left ( x^{2}-9 \right )\left ( {y}' \right )^{2}+x^{2}=0$
Hint:
You must know about the equation of circle
Given:
Family of circles having centre on y-axis and radius 3 units
Solution:
General equation of circle is
$\left ( x-a \right )^{2}+\left ( y-b \right )^{2}=r^{2}$
Given centre is on y-axis
$\therefore$ Centre $= \left ( 0,b \right )$ and Radius $= 3$
Hence, our equation is
$\left ( x-0 \right )^{2}+\left ( y-b \right )^{2}=3^{2}$
$x^{2}+\left ( y-b \right )^{2}=9$

Differentiate with respect to $x$ ,

$\begin{aligned} &2 x+2(y-b) y^{\prime}=0 \\ &2\left(x+(y-b) y^{\prime}\right)=0 \\ &(y-b) y^{\prime}=-x \\ &(y-b)=\frac{-x}{y^{\prime}} \end{aligned}$

Put value in (i)

$\begin{aligned} &x^{2}+(y-b)^{2}=9 \\ &x^{2}+\left[\frac{-x}{y^{\prime}}\right]^{2}=9 \\ &x^{2}+\frac{x^{2}}{y^{\prime 2}}=9 \\ &\frac{x^{2}\left(y^{\prime}\right)^{2}+x^{2}}{\left(y^{\prime}\right)^{2}}=9 \\ &x^{2}\left(y^{\prime}\right)^{2}+x^{2}=9\left(y^{\prime}\right)^{2} \\ &x^{2}\left(y^{\prime}\right)^{2}-9\left(y^{\prime}\right)^{2}+x^{2}=0 \\ &\left(y^{\prime}\right)^{2}\left[x^{2}-9\right]+x^{2}=0 \\ &\therefore \left(x^{2}-9\right)\left(y^{\prime}\right)^{2}+x^{2}=0 \end{aligned}$

Differential Equations Exercise Revision Exercise Question 8

Answer:
$xy{}'-2y=0$
Hint:
You must know about the parabolas to form an equation
Given:
Family of parabola having vertex at origin and axis along the positive y-direction
Solution:
$x^{2}=4ay$ … (i)
Where, $a$ is constant parameter
Differentiate with respect to $x$ ,
$\begin{aligned} &2 x=4 a \frac{d y}{d x} \\ &x=2 a y^{\prime} \\ &a=\frac{x}{2 y^{\prime}} \\ &\text { Put in (i) } \\ &x^{2}=4 \frac{x}{2 y^{\prime}} \cdot y \\ &y^{\prime}=\frac{2 y}{x} \\ &x y^{\prime}=2 y \\ &x y^{\prime}-2 y=0 \end{aligned}$

Differential Equations Exercise Revision Exercise Question 9

Answer:
$xy\frac{d^{2}y}{dx^{2}}+x\left ( \frac{dy}{dx} \right )^{2}-y\frac{dy}{dx}=0$
Hint:
You must know about the equation of ellipses
Given:
Family of ellipses having foci on y-axis and centre at origin
Solution:
Ellipse whose foci on y-axis
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ [Two constants, differentiate twice] $\begin{aligned} &\frac{d}{d x}\left[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right]=\frac{d}{d x}(1) \\ &\frac{1}{a^{2}}\left(\frac{d\left(x^{2}\right)}{d x}\right)+\frac{1}{b^{2}}\left(\frac{d\left(y^{2}\right)}{d x}\right)=0 \\ &\frac{1}{a^{2}}(2 x)+\frac{1}{b^{2}}\left(2 y \frac{d y}{d x}\right)=0 \\ &\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}}\left(\frac{d y}{d x}\right)=0 \\ &\frac{2 y}{b^{2}}\left(\frac{d y}{d x}\right)=-\frac{2 x}{a^{2}} \\ &\frac{y}{b^{2}}\left(\frac{d y}{d x}\right)=-\frac{x}{a^{2}} \\ &\frac{y}{x}\left(\frac{d y}{d x}\right)=-\frac{b^{2}}{a^{2}} \\ &\frac{y}{x} y^{\prime}=-\frac{b^{2}}{a^{2}} \end{aligned}$

Again differentiate,

Differential Equations Exercise Revision Exercise Question 10

Answer: $xy{y}''+x\left ( {y}' \right )^{2}-y{y}'=0$
Hint: You must know about the equation of hyperbola
Given: Family of hyperbola having foci on x-axis and centre at origin
Solution: Hyperbola whose foci on y-axis
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ [Two constants, differentiate twice]
$\begin{aligned} &\frac{d}{d x}\left[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\right]=\frac{d}{d x}(1) \\ &\frac{1}{a^{2}}\left(\frac{d\left(x^{2}\right)}{d x}\right)-\frac{1}{b^{2}}\left(\frac{d\left(y^{2}\right)}{d x}\right)=0 \\ &\frac{1}{a^{2}}(2 x)-\frac{1}{b^{2}}\left(2 y \frac{d y}{d x}\right)=0 \\ &\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}}\left(\frac{d y}{d x}\right)=0 \end{aligned}$ $\begin{aligned} &\frac{2 y}{b^{2}}\left(y^{\prime}\right)=\frac{2 x}{a^{2}} \\ &\frac{y}{b^{2}}\left(y^{\prime}\right)=\frac{x}{a^{2}} \\ &\frac{y}{x}\left(y^{\prime}\right)=\frac{b^{2}}{a^{2}} \\ &\frac{y}{x} y^{\prime}=\frac{b^{2}}{a^{2}} \end{aligned}$

Again differentiate,

$\begin{aligned} &y^{\prime} \frac{\left[\left(y y^{\prime}\right) x-y\right]}{x^{2}}+\frac{y}{x} y^{\prime \prime}=0 . . \text { using product and division rule of differentiation }\\ &\left(y^{\prime} y^{\prime}+y y^{\prime \prime}\right) x-y y^{\prime}=0 . . \text { multiplying by } x^{2}\\ &\left(y^{\prime 2}+y y^{\prime \prime}\right) x-y y^{\prime}=0\\ &x y y^{\prime \prime}+x\left(y^{\prime}\right)^{2}-y y^{\prime}=0 \end{aligned}$

Differential Equations Exercise Revision Exercise Question 11

Answer: $xy^{11}+2y^{1}-xy+x^{2}-2=0$
Hint: Find first and second derivative of equation put in given differential equation to verify.
Given: $x y=a e^{x}+b e^{-x}+x^{2}, \quad x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2 x$
Solution: $x y=a e^{x}+b e^{-x}+x^{2},$
$\mathrm{y}+\mathrm{x} \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\mathrm{ae}^{\mathrm{x}}-\mathrm{be}^{-\mathrm{x}}+2 \mathrm{x} \ldots \text { differentiating using product rule }$
$\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}+\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ae}^{\mathrm{x}}+\mathrm{be}^{-\mathrm{x}}+2 \ldots \text { differentiating again } w . r \cdot t \text { to } x \\$
$\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-2=\mathrm{ae}^{\mathrm{x}}+\mathrm{be}^{-\mathrm{x}} \\$
$\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-2=\mathrm{x} \mathrm{y}-\mathrm{x}^{2} \\$ $\left[\begin{array}{l} x y=a e^{x}+b e^{-x}+x^{2} \\ x y-x^{2}=a e^{x}+b e^{-x} \end{array}\right]$
$\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-\mathrm{x} \mathrm{y}+\mathrm{x}^{2}-2=0$
Hence proved

Differential Equations Exercise Revision Exercise Question 12

Answer: Hence verified
Hint: Find first and second derivative of given equation and put in differential equation to be verified.
Given: $y=cx+2c^{2}\; \; \; \; ,2\left ( \frac{dy}{dx} \right )^{2}+x\frac{dy}{dx}-y=0$
Solution: $y=C\: x+2\: c^{2}$
$\frac{dy}{dx}=c$ …differentiating w.r.t to x
$\frac{d^{2}y}{dx^{2}}=0$ ..differentiating agin w.r.t to $x$
Now, differential equation is
$\begin{aligned} &2\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+x \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0 \\ &\mathrm{~L} . \mathrm{H} \cdot \mathrm{S}=2(\mathrm{c})^{2}+\mathrm{x}(\mathrm{c})-\mathrm{c} x-2 \mathrm{c}^{2} \\ &=2 \mathrm{c}^{2}+\mathrm{c} \mathrm{x}-\mathrm{c} \mathrm{x}-2 \mathrm{c}^{2}=0=\mathrm{R} . \mathrm{H} . \mathrm{S} \end{aligned}$
Hence verified

Differential Equations Exercise Revision Exercise Question 13

Answer: $y^{2}-x^{2}-xy=a$ is the solution of the given differential equation
Hint: Find first and second order derivative and put values in differential equation to be verified.
Given: $y^{2}-x^{2}-xy=a,\left ( x-2y \right )\frac{dy}{dx}+2x+y=0$
Solution: $y^{2}-x^{2}-xy=a$
$\begin{aligned} &2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{x}-\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0 \text { differentiating } w \cdot r . t \text { to } x \\ &-\left[(x-2 y) \frac{d y}{d x}+2 x+y\right]=0 \\ &\therefore(\mathrm{x}-2 \mathrm{y}) \frac{d y}{d x}+2 x+y=0 \end{aligned}$
Thus $y^{2}-x^{2}-xy=a$ is the solution of the given differential equation

Differential Equations Exercise Revision Exercise Question 14

Answer: Thus : $y=A\cos x+\sin x$ is the solution of the given differential equation
Hint: Solve first order derivative and put values in differential equation to be verified.
Given: $y=A\cos x+\sin x,\cos x\frac{dy}{dx}+\left ( sinx \right )y=1$
Solution:
$\frac{dy}{dx}=A\sin x+\cos x$ …Differentiating w.r.t to x,
Now, differential eq.
$\begin{aligned} &\text { L. } H . S=\operatorname{cos} x\left(\frac{d y}{d x}\right)+(\operatorname{sin} x) y \\ &\operatorname{cos} x(-A \operatorname{sin} x+\operatorname{cos} x)+\operatorname{sin} x(A \operatorname{cos} x+\operatorname{sin} x) \\ &-A \operatorname{sin} x \operatorname{cos} x+\operatorname{cos}^{2} x+A \operatorname{sin} x \operatorname{cos} x+\operatorname{sin}^{2} x \\ &\operatorname{cos}^{2} x+\operatorname{sin}^{2} x=1=R . H . S \ldots\left\{\operatorname{cos}^{2} x+\operatorname{sin}^{2} x=1\right\} \end{aligned}$
Hence, Proved.

Differential Equations Exercise Revision Exercise Question 15

Answer: $\frac{d^{3} y}{d x^{8}}-7\left(\frac{d y}{d x}\right)+6 y=0$
Hint: The number constants is equals to the number time we differentiate.
Given: $y=ae^{2x}+be^{-3x}+ce^{x}$
Solution: $y=ae^{2x}+be^{-3x}+ce^{x}$
$\begin{aligned} &\frac{d y}{d x}=2 a e^{2 x}-3 b e^{-3 x}+c e^{x} \\ &\frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+9 b e^{-3 x}+c e^{x} \\ &\frac{d^{3} y}{d x^{3}}=8 a e^{2 x}-27 b e^{-3 x}+c e^{x} \\ &\frac{d^{3} y}{d x^{3}}=7\left(2 a e^{2 x}-3 b e^{-3 x}+c e^{x}\right)-6\left(a e^{2 x}+b e^{-3 x}+c e^{x}\right) \\ &\frac{d^{3} y}{d x^{3}}=7\left(\frac{d y}{d x}\right)-6 y \\ &\frac{d^{3} y}{d x^{3}}-7\left(\frac{d y}{d x}\right)+6 y=0 \end{aligned}$

Differential Equations Exercise Revision Exercise Question 16

Answer: $\frac{d^{3}y}{dx^{3}}=0$
Hint: You must know about the equation of all parabola
Given: equation of all parabolas which have their axes parallel to y-axis is $\frac{d^{3}y}{dx^{3}}=0$
Solution: Equation of the family of parabolas having axis parallel to y-axis is given by
$\left ( x-n \right )^{2}=4b\left ( y-k \right )$ ..where n,k are constants
Differentiating w.r.t to $x$ , $2\left ( x-n \right )=4b\frac{dy}{dx}$
$\left ( x-n \right )=2b\frac{dy}{dx}$
Differentiating again, $1=2b\left ( \frac{d^{2}y}{dx^{2}} \right )$
Again Differentiating, $0=2b\left ( \frac{d^{3}y}{dx^{3}} \right )$
$i.e.$ $\frac{d^{3}y}{dx^{3}}=0$

Differential Equations Exercise Revision Exercise Question 17

Answer: $3y^{{}'}\left ( y^{{}'''} \right )^{2}=y^{{}'''}\left ( 1+\left ( y^{{}'} \right )^{2} \right )$
Hint: The number of constant is equal to the number of times we need to differentiate.
Given: $x^{2}+y^{2}+2ax+2by+C=0$ , find differential equation, not containing constants.
Solution: $x^{2}+y^{2}+2ax+2by+C=0$
Differentiate w.r.t x,
$2x + 2yy^{{}'} + 2a + 2by^{{}'} = 0$
Again, differentiate w.r.t x
$2 + 2(y^{{}'})^{2} + 2yy^{{}''} + 2by^{{}''} = 0$
$1 + (y^{{}'})^{2} + yy^{{}''} + by^{{}''} = 0$ ..taking 2 common
$b=\frac{-\left (1+\left ( y^{{}'} \right )^{2}+yy^{{}''} \right )}{y^{{}''}}$
We have,
$1+\left(y^{{}'}\right)^{2}+y y^{{}''}+b y^{{}''}=0$
Again differentiate,
$2 y^{{}'} y^{{}''}+y^{{}'} y^{{}''}+y y^{{}'''}+b y^{{}'''}=0$
On substituting values of (b),
$\begin{aligned} &3 y^{{}'} y^{{}''}+y y^{{}'''}+\left(\frac{-\left(1+\left(y^{{}'}\right)^{2}+y y^{{}''}\right.}{y^{{}''}}\right) y^{{}'''}=0 \\ &3 y^{{}'}\left(y^{{}''}\right)^{2}+y y^{{}''} y^{{}'''}-y^{{}''}-\left(y^{{}'}\right)^{2} y^{{}'''}-y y^{{}'''} y^{{}''}=0 \\ &3 y^{{}'}\left(y^{{}''}\right)^{2}=y^{{}'''}\left(1+\left(y^{{}'}\right)^{2}\right) \end{aligned}$

Differential Equations Exercise Revision Exercise Question 18

Answer: $y=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+\frac{2(x+1)^{5 / 2}}{5}-\frac{2(x+1)^{8 / 2}}{3}+C$
Hint: Apply integration by parts method.
Given: $\frac{dy}{dx}=\sin ^{3}x\cos ^{4}x+x\sqrt{x+1}$
Solution: $\frac{dy}{dx}=\sin ^{3}x\cos ^{4}x+x\sqrt{x+1}$
$\begin{aligned} &\left.\Rightarrow \int d y=\int \sin ^{3} x \cos ^{4} x+x \sqrt{x+1}\right) d x \text { (integrating both sides) } \\ &\left.\Rightarrow \quad y=\int \sin ^{3} x \cos ^{4} x d x+\int x \sqrt{x+1}\right) d x \\ &\Rightarrow \quad y=I_{1}+I_{2} \end{aligned}$
Now, $I_{1}=\int \sin ^{3}x\cos ^{4}xdx$
$=\int \left ( 1-\cos ^{2}x \right )\left ( \cos ^{4}x \right )\left ( \sin x \right )dx\; \; \; \; \; \; \; \; \; \; \; \left ( \because \cos ^{2}x+\sin ^{2}x=1 \right )$
Let $t=\cos x$
$\begin{aligned} &\Rightarrow \quad d t=-\sin x d x(\text { diff. w.r.tx }) \\ &\Rightarrow I_{1}=-\int t^{4}\left(1-t^{2}\right) d t \\ &\left.\Rightarrow I_{1}=\int t^{4}\left(t^{2}-1\right) d t=\int t^{6}-t^{4}\right) d t \\ &\Rightarrow \quad l_{1}=\frac{t^{7}}{7}-\frac{t^{5}}{5}+c_{1} \\ &=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+c_{1} \\ &\left.l_{2}=\int x \sqrt{x+1}\right) d x \end{aligned}$
Let $t^{2}=x+1$
$\begin{aligned} &\Rightarrow 2 t d t=d x \quad(\text { diff. } w . r \cdot t x) \\ &\Rightarrow \quad l_{2}=2 \int\left(t^{2}-1\right) \cdot t \cdot t d t \\ &\Rightarrow \quad l_{2}=2 \int t^{4}-t^{2} d t \\ &\Rightarrow \quad I_{2}=\frac{2 t^{5}}{5}-\frac{2 t^{3}}{3}+c_{2} \end{aligned}$
$\therefore y=I_{1}+I_{2}$
$\begin{aligned} &\Rightarrow \mathrm{y}=\frac{\cos ^{7} \mathrm{x}}{7}-\frac{\cos ^{5} \mathrm{x}}{5}+\mathrm{c}_{1}+\frac{2(\mathrm{x}+1)^{5 / 2}}{5}-\frac{2(\mathrm{x}+1)^{8 / 2}}{3}+\mathrm{C}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{7} \mathrm{x}}{7}-\frac{\cos ^{5} \mathrm{x}}{5}+\frac{2(\mathrm{x}+1)^{5 / 2}}{5}-\frac{2(\mathrm{x}+1)^{8 / 2}}{3}+\mathrm{c} \quad\left(\because \mathrm{c}_{1}+\mathrm{C}_{2}=\mathrm{C}\right) \end{aligned}$

Differential Equations Exercise Revision Exercise Question 19

Answer: $y=\tan ^{-1}\left ( x+2 \right )+c$
Hint: Use the formula of $\int \frac{1}{x^{2}+1}dx$
Given: $\frac{dy}{dx}=\frac{1}{x^{2}+4x+5}$
Solution: $\frac{dy}{dx}=\frac{1}{x^{2}+4x+5}$ $\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{1}{x^{2}+4 x+4+1} \\ &\Rightarrow \frac{d y}{d x}=\frac{1}{(x+2)^{2}+(1)^{2}} \\ &\Rightarrow d y=\frac{1}{(x+2)^{2}+(1)^{2}} d x \end{aligned}$

Integrating both sides, we get

$\begin{aligned} &\int \mathrm{dy}=\int \frac{1}{(x+2)^{2}+(1)^{2}} \mathrm{dx} \\ &\Rightarrow \mathrm{y}=\tan ^{-1}\left(\frac{x+2}{1}\right)+\mathrm{c} \ldots \int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ &\Rightarrow \mathrm{y}=\tan ^{-1}(x+2)+\mathrm{c} \end{aligned}$

Differential Equations Exercise Revision Exercise Question 20

Answer: $x=\tan ^{-1}\left ( y+1 \right )+c$
Hint: Separate y & x and than integrate both sides
Given: $\frac{dy}{dx}=y^{2}+2y+2$
Solution: $\frac{dy}{dx}=y^{2}+2y+2$ $\begin{aligned} &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}+2 \mathrm{y}+1+1 \\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{y}+1)^{2}+(1)^{2} \\ &\Rightarrow \frac{1}{(\mathrm{y}+1)^{2}+(1)^{2}} \mathrm{dy}=\mathrm{dx} \end{aligned}$

Integrating both sides, we get

$\begin{aligned} &\Rightarrow \int \frac{1}{(\mathrm{y}+1)^{2}+(1)^{2}} \mathrm{dy}=\int \mathrm{dx} \ldots \int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ &\Rightarrow \tan ^{-1}\left(\frac{y+1}{1}\right)+\mathrm{c}=\mathrm{x} \\ &\Rightarrow \mathrm{x}=\tan ^{-1}(y+1)+\mathrm{c} \end{aligned}$

Differential Equations Exercise Revision Exercise Question 21

Answer: $y+2x^{2}=e^{x}+c$
Hint: Apply integration to find the equation
Given: $\frac{dy}{dx}+4x=e^{x}$
Solution: $\frac{dy}{dx}+4x=e^{x}$
$\begin{aligned} &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}-4 \mathrm{x} \\ &\Rightarrow \mathrm{dy}=\left(\mathrm{e}^{\mathrm{x}}-4 \mathrm{x}\right) \mathrm{dx} \end{aligned}$
integrate both sides
$\begin{aligned} &\Rightarrow \int d y=\int\left(e^{x}-4 x\right) d x \\ &\Rightarrow \quad y=e^{x}-4 * \frac{x^{2}}{2}+c \\ &\Rightarrow \quad y=e^{x}-2 * x^{2}+c \\ &\Rightarrow \quad y+2 x^{2}=e^{x}+c \end{aligned}$

Differential Equations Exercise Revision Exercise Question 22

Answer: $y=\left ( x^{2}-2x +2\right )e^{x}+c$
Hint: You must know about the formula of $\int uv\; \; dx$
Given: $\frac{dy}{dx}=x^{2}e^{x}$
Solution: $\frac{dy}{dx}=x^{2}e^{x}$
$\begin{aligned} &\Rightarrow d y=x^{2} e^{x} d x\\ &\Rightarrow \int d y=\int x^{2} e^{x} d x \text { (integrate both sides) }\\ &\Rightarrow \int d y=x^{2} \int e^{x} d x-\int\left(2 x \int e^{x} d x\right) d x\\ &\left[\because \int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right]\\ &\Rightarrow y=x^{2} e^{x}-2 \int x e^{x} d x\\ &\Rightarrow y=x^{2} e^{x}-2 x \int e^{x} d x+2 \int e^{x} d x\\ &\left[\because \int \mathrm{uv} \mathrm{d} \mathrm{x}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}\right) \mathrm{dx}\right]\\ &\Rightarrow y=x^{2} e^{x}-2 x e^{x}+2 e^{x}+C\\ &\Rightarrow y=\left(x^{2}-2 x+2\right) e^{x}+C \end{aligned}$

Differential Equations Exercise Revision Exercise Question 23

Answer: $y=log|logx|+\frac{x^{2}}{2}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+c$
Hint: integrate both sides and then apply the formula of
Given: $\frac{dy}{dx}-x\sin ^{2}x=\frac{1}{xlogx}$
Solution: $\frac{dy}{dx}-x\sin ^{2}x=\frac{1}{xlogx}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x} \log x}+\mathrm{x} \sin ^{2} \mathrm{x}$
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{1}{\mathrm{x} \operatorname{logx}}+\frac{\mathrm{x}}{2}(1-\cos 2 \mathrm{x})\left[\because \operatorname{Cos} 2 \mathrm{x}=1-2 \sin ^{2} \mathrm{x}\right]$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x} \log x}+\frac{\mathrm{x}}{2}-\frac{x \operatorname{Cos} 2 \mathrm{x}}{2} \\$
$\Rightarrow \int \mathrm{dy}=\int\left(\frac{1}{\mathrm{xlog} \mathrm{x}}+\frac{\mathrm{x}}{2}-\frac{x \operatorname{Cos} 2 \mathrm{x}}{2}\right) \mathrm{d} \mathrm{x} \quad \text { (integrate both sides) } \\$
$\Rightarrow \int \mathrm{dy}=\int\left(\frac{1}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}+\frac{1}{2} \int \mathrm{x} \mathrm{dx}-\frac{1}{2}(\mathrm{x} \operatorname{Cos} 2 \mathrm{x}) \mathrm{d} \mathrm{x}\right. \\$
$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}}{2} \int \cos 2 \mathrm{x} \mathrm{dx}+\frac{1}{2} \int \frac{\sin 2 \mathrm{x}}{2} \mathrm{dx} \\$
$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}}{2} \cdot \frac{\sin 2 \mathrm{x}}{2}+\frac{1}{2}\left(-\frac{\cos 2 \mathrm{x}}{4}\right)+\mathrm{C} \\$
$\Rightarrow \mathrm{y}=\log |\log \mathrm{x}|+\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x} \sin 2 \mathrm{x}}{4}-\frac{\cos 2 \mathrm{x}}{8}+\mathrm{C}$

Differential Equations Exercise Revision Exercise Question 24

Answer: $y=log|\tan ^{2}+2\tan x+5|+C$
Hint: Use substitution method
Given: $\left(\tan ^{2} x+2 \tan x+5\right) \frac{d y}{d x}=2(1+\tan x) \sec ^{2} x$
Solution: $\left(\tan ^{2} x+2 \tan x+5\right) \frac{d y}{d x}=2(1+\tan x) \sec ^{2} x$
$\Rightarrow \quad d y=\frac{\left.2(1+\tan x) \sec ^{2} x\right)}{\left(\tan ^{2} x+2 \tan x+5\right)} d x$
Integrating both sides, we get,
$\Rightarrow \quad\int d y=\frac{\left.2(1+\tan x) \sec ^{2} x\right)}{\left(\tan ^{2} x+2 \tan x+5\right)} d x$
Let, $\tan ^{2}x+2\tan x+5=t$
Differentiate w.r.t x
$\begin{aligned} &\left(2 \tan x \sec ^{2} x+2 \sec ^{2} x\right) d x=d t \\ &\Rightarrow 2(1+\tan x) \sec ^{2} x d x=d t \\ &\Rightarrow \int d y=\int \frac{1}{t} d t \\ &\Rightarrow y=\log |t|+C \\ &\Rightarrow y=\log \left|\tan ^{2}+2 \tan x+5\right|+C \end{aligned}$

Differential Equations Exercise Revision Exercise Question 25

Answer: $\frac{\cos ^{5}}{5}-\frac{\cos ^{3}x}{3}+\left ( x-1 \right )e^{x}+C$
Hint: Apply integration by parts method and formula of $\int u\; v\: dx$
Given: $\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}$
Solution: $\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}$
$\Rightarrow \int dy=\int \left ( \sin ^{3}x\cos ^{2}x+xe^{x} \right )dx$ (integrate both sides)
$\Rightarrow \int y=\int \sin ^{3}x\cos ^{2}xdx+\int xe^{x}dx$
$\Rightarrow y=I_{1}+I_{2}$
$I_{1}=\int \sin ^{3}x\cos ^{2}xdx$
$=\int \left ( 1-\cos ^{2}x \right )\cos ^{2}x\sin xdx$
Let $\cos x=t$
$\Rightarrow d \mathrm{t}=-\sin \mathrm{x} \mathrm{d} \mathrm{x} \quad \text { (diff. w.r.t. } \mathrm{x})$
$I_{1}=-\int t^{2}\left ( 1-t^{2} \right )dt$
$=\int \mathrm{t}^{2}\left(\mathrm{t}^{2}-1\right) \mathrm{dt}=\int \mathrm{t}^{4}-\mathrm{t}^{2} \mathrm{dt} \\$
$\Rightarrow \mathrm{I}_{1}=\frac{\mathrm{t}^{5}}{5}-\frac{\mathrm{t}^{8}}{3}+\mathrm{c}_{1} \\$
$\Rightarrow \mathrm{I}_{1}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}$
Now, $I_{2}=\int xe^{x}dx$
$\begin{aligned} &\Rightarrow \mathrm{I}_{2}=x \int \mathrm{e}^{x} \mathrm{dx}-\int \mathrm{e}^{\mathrm{x}} \mathrm{dx} \\ &{\left[\because \int \mathrm{uvd} \mathrm{x}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}\right) \mathrm{dx}\right]} \\ &\Rightarrow \mathrm{I}_{2}=x \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\mathrm{c}_{2} \\ &\text { Now, } \mathrm{y}=\mathrm{I}_{1}+\mathrm{I}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}\left(\because \mathrm{c}_{1}+\mathrm{c}_{2}=\mathrm{c}\right) \end{aligned}$

Differential Equations Exercise Revision Exercise Question 26

Answer: $\sin x\sin y=c$
Hint: You must know about trigonometric identities.
Given: $\tan y\; dx+\tan \: x\; dy=0$
Solution: $\tan y\; dx+\tan \: x\; dy=0$
$\begin{aligned} &\Rightarrow \tan x d y=-\tan y d x \\ &\Rightarrow \frac{1}{-\tan y} d y=\frac{1}{\tan x} d x \\ &\Rightarrow \cot y d y=-\cot x d x\left(\because \frac{1}{\tan x}=\cot x\right) \end{aligned}$
Integrating both sides
$\begin{aligned} &\Rightarrow \int \cot y d y=-\int \cot x d x \\ &\Rightarrow \log |\sin y|=-\log |\sin x|+\log c \\ &\Rightarrow \log |\sin y|+\log |\sin x|=\log c \\ &\Rightarrow \log |\sin y \sin x|=\log c(\because \log a+\log b=\log a b) \\ &\Rightarrow \sin y \sin x=c \end{aligned}$

Differential Equations Exercise Revision Exercise Question 27

Answer: $x+y+log|xy|=c$
Hint: Separate x & y and then integrate
Given: $\left ( 1+x \right )y\: dx+\left ( 1+y \right )x\: dy=0$
Solution: $\left ( 1+x \right )y\: dx+\left ( 1+y \right )x\: dy=0$
$\begin{aligned} &\Rightarrow \quad\left(\frac{1+x}{x}\right) d x=-\left(\frac{1+y}{y}\right) d y \\ &\Rightarrow \quad\left(\frac{1}{y}+1\right) d y=-\left(\frac{1}{x}+1\right) d x \end{aligned}$
Integrating both sides, we get
$\begin{aligned} &\int\left(\frac{1}{y}+1\right) d y=-\int\left(\frac{1}{x}+1\right) d x \\ &\Rightarrow \int \frac{1}{y} d y+\int d y=-\int \frac{1}{x} d x-\int d x \\ &\Rightarrow \log |y|+y=-\log |x|-x+c \\ &\Rightarrow \log |y|+\log |x|+x+y=c \\ &\Rightarrow x+y+\log |x y|=c(\because \log a+\log b=\log a b) \end{aligned}$

Differential Equations Exercise Revision Exercise Question 28

Answer: $x\tan x-y\tan |\frac{\sec x}{\sec y}|+C$
Hint: Use the formula of $\int u\: v\: dx$
Given: $x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy$
Solution: $x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy$
$\begin{aligned} &\Rightarrow \quad \frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y \\ &\Rightarrow \quad y \sec ^{2} y d y=x \sec ^{2} x d x \quad\left(\because \frac{1}{\cos x}=\sec x\right) \\ &\Rightarrow \int y \sec ^{2} y d y=\int x \sec ^{2} x d x \quad(\because \text { integrating both sides }) \\ &\Rightarrow \quad y \int \sec ^{2} y d y-\int \tan y d y=x \int \sec ^{2} x d x-\int \tan x d x \\ &\quad\left[\because \int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right. \\ &\Rightarrow \quad y \tan y-\log |\sec y|=x \tan x-\log |\sec x|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log \left|\frac{\sec x}{\sec y}\right|+C \quad[\because \log a-\log b=\log (a / b)] \end{aligned}$

Differential Equations Exercise Revision Exercise Question 29

Answer: $\left [ log\left ( sec\: y+\tan \: y \right ) \right ]^{2}=\left [ log\left ( \sec \: x+\tan \: x \right ) \right ]^{2}+C$
Hint: Apply substitution method.
Given: $\cos \: y\: log|\sec \: x+\tan \: x|dx=\cos \: x\: log|\sec \: y+\tan \: y|dy$
Solution: $\cos \: y\: log|\sec \: x+\tan \: x|dx=\cos \: x\: log|\sec \: y+\tan \: y|dy$
$\begin{aligned} &\Rightarrow \frac{\log |\sec x+\tan x|}{\cos x} d x=\frac{\log |\sec y+\tan y|}{\cos y} d y\\ &\text { Integrating both sides }\\ &\Rightarrow \int \frac{\log |\sec y+\tan y|}{\cos y} d y=\int \frac{\log |\sec x+\tan x|}{\operatorname{Cos} x} d x \end{aligned}$
Let $\log |\sec y+\tan y|=t \text { and } \log |\sec x+\tan x|=u \\$
$\Rightarrow\left(\frac{\sec ^{2} y+s e c y \text { tany }}{s e c y+\tan y}\right) d y=d t \text { and }\left(\frac{\sec ^{2} x+\sec x \tan x}{\sec x+\tan x}\right) d x=d u \\$
$\Rightarrow \sec y d y=d t \quad \text { and } \sec x d x=d u \\$
$\Rightarrow\left(\frac{1}{c o s y}\right) d y=d t \text { and }\left(\frac{1}{\cos x}\right) d x=d u \\ \$
$\Rightarrow \int\: t\: d t=\int u d u \\$
$\Rightarrow \frac{t^{2}}{2}=\frac{u^{2}}{2}+c_{1} \\$
$\Rightarrow t^{2}=u^{2}+2 c_{1} \\$
$\Rightarrow t^{2}=u^{2}+c\left(\because 2 c_{1}=C\right) \\$
$\Rightarrow[\log (\sec y+\tan y)]^{2}=[\log (\sec x+\tan x)]^{2}+c$

Differential Equations Exercise Revision Exercise Question 30

$\Rightarrow$ Answer: $\frac{1}{2}\left ( log\: y \right )^{2}+\left ( 2-x^{2} \right )\cos x+2x\sin x=C$
Hint: Apply integration and then the formula of $\int u\: v\: dx$ and substitution method.
Given: $\cos cx\left ( log\: y \right )dy+x^{2}y \: d\: x=0$
Solution: $\cos cx\left ( log\: y \right )dy+x^{2}y \: d\: x=0$
$\begin{aligned} &\Rightarrow \frac{\log y}{y} d y=-\frac{x^{2}}{\operatorname{cosec} x} d x \\ &\Rightarrow \int \frac{l o g y}{y} d y=-\int x^{2} \sin x d x \end{aligned}$ (integrating both sides)
Let log y=t
$\begin{aligned} &\Rightarrow \frac{1}{y} d y=d t \quad(\text { diff. w.r.t. } y) \\ &\Rightarrow \because \int t d t=-\int x^{2} \sin x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=-x^{2}(-\cos x)+\int 2 x(-\cos x) d x \\ &\because\left[\int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right] \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 \int x \cos x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 x \sin x+2 \int \sin x d x \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=x^{2} \cos x-2 x \sin x-2 \cos x+C \\ &\Rightarrow \frac{1}{2}(\log y)^{2}=\left(x^{2}-2\right) \cos x-2 x \sin x+C \\ &\Rightarrow \frac{1}{2}(\log y)^{2}+\left(2-x^{2}\right) \cos x+2 x \operatorname{Sin} x=C \end{aligned}$

Differential Equations Exercise Revision Exercise Question 31

Answer: $log|y|+log|y-1|=-\frac{1}{2}log|1-x^{2}|+C$
Hint: you must know the rules of solving differential equation. First rearrange the values and then solve.
Given: $\left ( 1-x^{2} \right )dy+xy\: d\: x=xy^{2}dx$
Solution: $\left ( 1-x^{2} \right )dy+xy\: d\: x=xy^{2}dx$
$\begin{aligned} &\left(1-x^{2}\right) d y=\left(x y^{2}-x y\right) d x \\ &\left(1-x^{2}\right) d y=x y(y-1) d x \\ &\frac{1}{y(y-1)} d y=\frac{x}{1-x^{2}} d x \end{aligned}$
Integrating both sides, we get,
$\begin{aligned} &\int \frac{1}{\mathrm{y}(\mathrm{y}-1)} \mathrm{dy}=\int \frac{\mathrm{x}}{1-\mathrm{x}^{2}} \mathrm{dx} \ldots . .(\mathrm{I}) \\ &\text { L.H.S }:-\frac{1}{\mathrm{y}(\mathrm{y}-1)}=\frac{\mathrm{A}}{\mathrm{y}}+\frac{\mathrm{B}}{\mathrm{y}-1} \\ &1=\mathrm{A}(\mathrm{y}-1)+\mathrm{By} \end{aligned}$
Substituting $y=1$
$1=A\left ( 0-1 \right )+B\left ( 0 \right )$
$1=B$
Again substituting, $y=0$
$1=A\left ( 0-1 \right )+B(0)$
$1=-A$
$A=-1$
Substituting values of A and B in $\frac{1}{y\left ( y-1 \right )}=\frac{A}{y}+\frac{B}{y-1}$ , we get,
$\begin{aligned} &\frac{1}{y(y-1)}=\frac{-1}{y}+\frac{1}{y-1} \\ &\int \frac{1}{y(y-1)} d y=\int \frac{-1}{y} d y+\int \frac{1}{y-1} d y \\ &=-\log |y|+\log |y-1|+C_{1} \end{aligned}$
Now considering R.H.S of (II), we have,
$\int \frac{x}{1-x^{2}}dx$
Here, putting $1-x^{2}=t$ and differentiate both sides, we get,
$\begin{aligned} &-2 x d x=d t \\ &x d x=\frac{-d t}{2} \\ &\because \int \frac{x}{1-x^{2}} d x \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{-1}{2} \int \frac{1}{t} d t \\ &\Rightarrow \frac{-1}{2} \log |t|+C_{2} \quad\left[\because \int \frac{d t}{t}=\log |t|+C\right] \\ &\Rightarrow \frac{-1}{2} \log \left|1-x^{2}\right|+C_{2}\left[\text { where } t=1-x^{2}\right] \end{aligned}$
Now substituting the values of $\int \frac{1}{y\left ( y-1 \right )}dy$ and $\int \frac{x}{1-x^{2}}dx$ in (I)
$\begin{aligned} &\Rightarrow-\log |y|+\log |y-1|+C_{1}=-\frac{1}{2} \log \left|1-x^{2}\right|+c_{2} \\ &\Rightarrow-\log |y|+\log |y-1|=-\frac{1}{2} \log \left|1-x^{2}\right|+C \quad\left[\because \text { where } C_{2}-C 1=C\right] \end{aligned}$

Differential Equations Exercise Revision Exercise Question 32

Answer: $y^{2}log\: y=x\: \sin +C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}=\frac{\sin x+x\cos x}{y\left ( 2logy+1 \right )}$
Solution: $\frac{dy}{dx}=\frac{\sin x+x\cos x}{y\left ( 2logy+1 \right )}$
$y\left ( 2log\: y+1 \right )dy=\left ( \sin x+x\cos x \right )dx$
integrating both sides,
$\int 2\: y\: log \: y dy+1\int y \: dy=\int \sin x\: dx+\int x\cos x\: dx$
Integrating by parts,
$\begin{aligned} &2 \log y \int y d y-2 \int\left(\frac{d}{d y}(\log y) x \int y d y\right) d y+\int y d y=-\cos x+x \int \cos x d x-\int \frac{d}{d x}(x) \cdot \int \cos x d x \\ &\quad \Rightarrow 2 \log y\left(\frac{y^{2}}{2}\right)-\int y d y+\int y d y=-\cos x+x \sin x+\cos x+c \\ &\quad \Rightarrow y^{2} \log y=x \sin x+c \end{aligned}$

Differential Equations Exercise Revision Exercise Question 33

Answer: $\mathrm{e}^{y}+\mathrm{e}^{-y}-\log |\mathrm{x}|+\frac{\mathrm{x}^{2}}{2}+\mathrm{C}$
Hint: you must know the rules of solving differential equation and integrations.
Given: $x\left ( e^{2y}-1 \right )dy+\left ( x^{2}-1 \right )e^{y}dx=0$
Solution: $x\left ( e^{2y}-1 \right )dy+\left ( x^{2}-1 \right )e^{y}dx=0$
$\begin{aligned} &x\left(e^{2 y}-1\right) d y=-\left(x^{2}-1\right) e^{y} d x \\ &x\left(e^{2 y}-1\right) d y=\left(1-x^{2}\right) e^{y} d x \\ &\frac{e^{2 y}-1}{e^{y}} d y=\frac{1-x^{2}}{x} d x \\ &\left(\frac{e^{2 y}}{e^{y}}-\frac{1}{e^{y}}\right) d y=\left(\frac{1}{x}-\frac{x^{2}}{x}\right) d x \\ &\left(e^{2 y-y}-e^{-y}\right) d y=\left(\frac{1}{x}-x\right) d x \end{aligned}$
Integrating both sides,
$\int\left(e^{y}-e^{-y}\right) d y=\int \frac{1}{x} d x-\int x d x \\ \int\left(e^{y} d y-\int e^{-y} d y=\int \frac{1}{x} d x-\int x d x\right. \\$
$e^{y}-\left(-e^{-y}\right)=\log |x|-\frac{x^{2}}{2}+c$
$\Rightarrow e^{y}+e^{-y}-\log |x|+\frac{x^{2}}{2}+c$

Differential Equations Exercise Revision Exercise Question 34

Answer: $\left ( x+c \right )e^{x+y}+1=0$ \

Hint: you must know the rules of solving differential equation and integrations.

Given: $\frac{dy}{dx}+1=e^{x+y}$

Solution: $\frac{dy}{dx}+1=e^{x+y}$ ..........(I)

Put x + y = t and differentiate both sides. We get,

$1+\frac{dy}{dx}=\frac{dt}{dx}$

Compare with equation (I),

$\frac{dt}{dx}=e^{t}$

$e^{-t}dt=dx$

Now, integrating both sides,

$\begin{aligned} &\int \mathrm{e}^{-\mathrm{t}} \mathrm{d} \mathrm{t}=\int \mathrm{dx} \quad\left[\because \int \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\frac{\mathrm{e}^{-\mathrm{t}}}{-1}\right] \\ &-\mathrm{e}^{-\mathrm{t}}=\mathrm{x}+\mathrm{C} \\ &\frac{-1}{\mathrm{e}^{\mathrm{x}+\mathrm{y}}}=\mathrm{x}+\mathrm{C} \\ &(\mathrm{x}+\mathrm{C}) \mathrm{e}^{\mathrm{x}+\mathrm{y}}+1=0 \end{aligned}$

Differential Equations Exercise Revision Exercise Question 35

Answer: $\tan^{-1}\left ( x+y \right )=x+c$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}=\left ( x+y \right )^{2}$
Solution: $\frac{dy}{dx}=\left ( x+y \right )^{2}$
Let $\left ( x+y \right )=u$ and differentiate both sides ,we get,
$\begin{aligned} &1+\frac{d y}{d x}=\frac{d u}{d x} \\ &\frac{d y}{d x}=\frac{d u}{d x}-1 \quad(x+y)=u \quad\left[\because(x+y)^{2}=u^{2]}\right. \\ &U^{2}=\frac{d u}{d x}-1 \quad\left[\because \frac{d u}{d x}=u^{2}\right] \\ &\frac{d u}{d x}=u^{2}+1 \\ &\frac{d u}{u 2+1}=d x \end{aligned}$
Now, integrating both sides,
$\begin{aligned} &\int \frac{\mathrm{du}}{\mathrm{u} 2+1}=\int \mathrm{dx} \\ &\tan ^{-1}=\mathrm{x}+\mathrm{C} \quad\left[\int \frac{1}{x^{2}+1} d x=\tan ^{-1} \mathrm{x}+\mathrm{C}\right] \\ &\therefore \tan ^{-1}(\mathrm{x}+\mathrm{y})=\mathrm{x}+\mathrm{c} \text { where, } \mathrm{u}=\mathrm{x}+\mathrm{y} \end{aligned}$
Hence, $\tan ^{-1}\left ( x+y \right )=x+c$

Differential Equations Exercise Revision Exercise Question 36

Answer: $y=\tan \left ( \frac{x+y}{2} \right )+C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\cos \left ( x+y \right )dy=dx$
Solution: $\cos \left ( x+y \right )dy=dx$ $\left ( \frac{dy}{dx}=\frac{1}{\cos \left ( x+y \right )} \right )$
Let $\left ( x+y \right )=u$ and differentiating both sides,
$\begin{aligned} &1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}-1 \\ &\frac{1}{\cos \mathrm{u}}=\left(\frac{\mathrm{du}}{\mathrm{dx}}-1\right) \\ &1=\left(\frac{\mathrm{du}}{\mathrm{dx}}-1\right) \cdot \cos \mathrm{u} \\ &\operatorname{Cos} \mathrm{u} \frac{\mathrm{du}}{\mathrm{dx}}=1+\cos \mathrm{u} \end{aligned}$
$\begin{aligned} &\frac{\text { cosu }}{1+\operatorname{cosu}} d u=d x \\ &\frac{(1+\cos u)-1}{1+\cos u} d u=d x \\ &{\left[\frac{1+\cos u}{1+\operatorname{cosu}}-\frac{1}{1+\operatorname{cosu}}\right] d u=d x} \\ &{\left[1-\frac{1}{1+\operatorname{cosu}}\right] d u=d x} \end{aligned}$
$\begin{aligned} &\Rightarrow\left[1-\frac{1}{2 \cos ^{2} \frac{u}{2}}\right] d u=d x \quad\left[\because \frac{1}{1+\cos x}=2 \cos ^{2} \frac{u}{2}\right. \\ &\Rightarrow\left[1-\frac{1}{2} \sec ^{2} \frac{u}{2}\right] d u=d x \end{aligned}$
Now, integrating both sides,
$\begin{aligned} &\int 1 \mathrm{du}=\int 1 .-\frac{1}{2} \sec ^{2} \frac{u}{2} \mathrm{du}=\int \mathrm{dx} \\ &\mathrm{U}-\tan \frac{\mathrm{u}}{2}=\mathrm{x}+\mathrm{c} \quad\left[\because \int \sec ^{2} \frac{\mathrm{x}}{2}=2 \tan \frac{\mathrm{x}}{2}\right] \end{aligned}$
Put the value of u
$\begin{aligned} &(x+y)-\tan \frac{(x+y)}{2}=x+c \quad[\because u=x+y] \\ &\therefore y=\tan \left(\frac{x+y}{2}\right)+C \end{aligned}$

Differential Equations Exercise Revision Exercise Question 37

Answer:
$y-2x=cx^{2}y$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$
Solution: $\frac{dy}{dx}+\frac{y}{x}=\frac{y^{2}}{x^{2}}$
$\frac{dy}{dx}=\frac{y^{2}}{x^{2}}-\frac{y}{x}$
$\frac{dy}{dx}=\frac{y^{2}xy}{x^{2}}$
Put y = v x and differentiate both sides w.r.t x
$\begin{aligned} &\frac{d y}{d x}=v .1+x \frac{d v}{d x} \\ &\frac{d y}{d x}=v+x \frac{d v}{d x} \end{aligned}$
So, equation (I) becomes
$\begin{aligned} &\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}^{2}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}^{2}-\mathrm{v}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{d} \mathrm{x}}=\mathrm{v}^{2}-2 \mathrm{v} \\ &\left(\frac{1}{\mathrm{v}^{2} 2 \mathrm{v}}\right) \mathrm{d} \mathrm{v}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$
Now, integrating both sides
$\begin{aligned} &\int\left(\frac{1}{(v-1)^{2}-1} d v=\int \frac{1}{x} d x\left[\because(v-1)^{2}-1=v^{2}+1-2 v-1=v^{2}-2 v\right]\right. \\ &\frac{1}{2} \log \left(\frac{v-2}{v}\right)=\log |x|+\log c \\ &\frac{1}{2} \log \left(\frac{v-2}{v}\right)=\log c x \\ &\log \left(\frac{v-2}{v}\right)=2 \log c x \\ &\log \left(\frac{v-2}{v}\right)=\log c^{2} x^{2}\left[\because \operatorname{alog} x=\log x^{a}\right] \end{aligned}$
$\begin{aligned} &\frac{\mathrm{v}-2}{\mathrm{v}}=\mathrm{c}^{2} \mathrm{x}^{2}\\ &1-\frac{2}{v}=c x^{2}[c=c]\\ &\left.1-\frac{2}{\frac{y}{x}}=c x^{2} \text { (put value of } v=\frac{y}{x}\right)\\ &1-\frac{2 x}{y}=c x^{2}\\ &y-2 x=c x^{2} y \end{aligned}$

Differential Equations Exercise Revision Exercise Question 38

Answer: $2A=log\left ( xy \right )-\frac{x}{y}$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}=\frac{y\left ( x-y \right )}{x\left ( x+y \right )}$
Solution: $\frac{dy}{dx}=\frac{y\left ( x-y \right )}{x\left ( x+y \right )}$
$x\left ( x+y \right )dy=y\left ( x-y \right )dx$ .....................(1)
Put , y=vx and differentiate both side.
$\frac{dy}{dx}=x\frac{dv}{dx}+v$
Eq. (1) becomes,
$\begin{aligned} &x(x+y x)(x d x+y d x)=y x(x-y x) d x \\ &x^{2}(1+v)(x d v+y d x)-v x^{2}(1-v) d x=0 \\ &x^{2}[(1+v) x d v+v(1+v) d x-v(1-v) d x]=0 \\ &x(1+v) d v+v(1+v) d x-v(1-v) d x=0 \\ &{[y(1+v) d x-v(1-v) d x]=-x(1+v) d v} \\ &{\left[v+v^{2}-v+v^{2}\right] d x=-x(1+v) d v} \\ &2 v^{2} d x=-x(1+v) d v \\ &\frac{d x}{-x}=\frac{1}{2 v^{2}}+\frac{v}{2 v^{2}} \int d v \\ &\frac{d x}{-x}=\frac{1}{2} v^{-2}+\frac{11}{2 v} \int d v \end{aligned}$
Now, integrating both sides,
$\begin{aligned} &\int \frac{\mathrm{dx}}{-\mathrm{x}}=\frac{1}{2} \int \mathrm{v}^{-2} \mathrm{dv}+\frac{1}{2} \int \frac{1}{\mathrm{v}} \mathrm{dv} \\ &-\log |\mathrm{x}|+\mathrm{A}=\frac{1}{2} \frac{\mathrm{v}^{-2+1}}{-2+1}+\frac{1}{2} \log |\mathrm{v}| \\ &-\log |\mathrm{x}|+\mathrm{A}=-\frac{1}{2} \mathrm{v}^{-1}+\frac{1}{2} \log \mathrm{v} \\ &-\log |\mathrm{x}|+\mathrm{A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}^{\frac{1}{2}} \\ &\mathrm{~A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}^{\frac{1}{2}}+\log |\mathrm{x}| \\ &\mathrm{A}=\frac{-1}{2 \mathrm{v}}+\log \mathrm{v}_{2} x \end{aligned}$
Put value of $v=\frac{y}{x}$

$\begin{aligned} &\mathrm{A}=\frac{-1}{2\left(\frac{y}{\mathrm{x}}\right)}+\log (\mathrm{xy})^{\frac{1}{2}} \\ &\mathrm{~A}=\frac{-\mathrm{x}}{2 \mathrm{y}}+\frac{1}{2} \log \mathrm{xy} \\ &2 \mathrm{~A}=\log (x y)-\frac{\mathrm{x}}{\mathrm{y}} \end{aligned}$ (where A is integration constant)

Differential Equations Exercise Revision Exercise Question 39

Answer:
$2\left ( x+y \right )-4x-log|2x+2y-1|=4C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\left ( x+y-1 \right )dy=\left ( x+y \right )dx$
Solution:
$\left ( x+y-1 \right )dy=\left ( x+y \right )dx$
$\frac{dy}{dx}=\frac{\left ( x+y \right )}{x+y-1}$
Putting $x+y=v$ , we get,
$\begin{aligned} &1+\frac{d y}{d x}=\frac{d v}{d x} \\ &\frac{d y}{d x}=\frac{d v}{d x}-1 \\ &\frac{d v}{d x}-1=\frac{v}{(v-1)} \\ &\frac{d v}{d x}=\frac{v}{(v-1)}+1 \\ &\frac{d v}{d x}=\frac{v+v-1}{(v-1)} \\ &\frac{d v}{d x}=\frac{2 v-1}{(v-1)} \\ &\frac{v-1}{2 v-1} d v=d x \end{aligned}$
Integration both sides we get
$\begin{aligned} &\int \frac{v-1}{2 v+1} d v=\int d x \\ &\frac{1}{2} \int \frac{2 v}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int \frac{2 v-1+1}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int d v+\frac{1}{2} \int \frac{1}{2 v-1} d v-\int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} \int d v-\frac{1}{2} \int \frac{1}{2 v-1} d v=\int d x \\ &\frac{1}{2} v-\frac{1}{4} \log |2 v-1|=x+C \end{aligned}$
$\begin{aligned} &\frac{1}{2}(x+y)-\frac{1}{4} \log |2 x+2 y-1|=x+C \\ &2(x+y)-\log |2 x+2 y-1|=4 x+4 C \\ &2(x+y)-4 x-\log |2 x+2 y-1|=4 C \end{aligned}$

Differential Equations Exercise Revision Exercise Question 40

Answer: $y\, \cos ec\: x+\cot \: x=c$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}-y\cot x=\cos ec\: x$
Solution: $\frac{dy}{dx}-y\cot x=\cos ec\: x$
The above equation is in form of $\frac{dy}{dx}+p\: y=q$

Where p = -cot x and q = cosec x

Integrating factor = $e^{\int px}$

Considering $\int p\: d\: x$

$\begin{aligned} \Rightarrow & \int p d x=-\int \cot x d x \\ &=-\log |\sin x| &\left[\because e^{\log x}=x\right] \\ & \therefore \mathrm{e}^{\int p d x}=e^{-\log |\sin x|}=e^{\log (\sin x)^{-1}} \\ &=\sin x^{-1}=\frac{1}{\sin x}=\operatorname{cosec} x \end{aligned}$

$\therefore \text { Integrating factor I.F }=\operatorname{cosec} \mathrm{x}$

Now, General solution is,

$\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int \mathrm{q}(\mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x} \operatorname{cosec} \mathrm{x} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x}^{2} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=-\cot \mathrm{x}+\mathrm{C} \\ &\Rightarrow \mathrm{y} \operatorname{cosec} x+\cot \mathrm{x}=\mathrm{c} \end{aligned}$

Differential Equations Exercise Revision Exercise Question 41

Answer: $y\cos x=\frac{\cos 2x}{2}+c$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}-y\tan x=-2\sin x$
Solution: $\frac{dy}{dx}-y\tan x=-2\sin x$
$\frac{dy}{dx}+\left ( -\tan x \right )y=-2\sin x$
The above equation look like,
$\frac{dy}{dx}+py=q$
Where p = -tan x and q = -2 sin x
Integrating factor = $e^{\int px}$
$=e^{\int -\tan x\: dx}$
$=e-^{\int \tan x\: dx}$
we have $\int \tan x\: dx=log\left ( \sec x \right )+c$ $\begin{aligned} &\text { I. } F=e^{-\log (s e c x)} \\ &\begin{array}{l} \text { I.F }=e^{-\log \left(\frac{1}{\operatorname{secx}}\right)} \\ \text { I.F }=\frac{1}{\sec x} \quad\left[\because \mathrm{e}^{\log x}=\mathrm{x}\right] \\ \therefore \text { I. } F=\operatorname{Cos} \mathrm{x} \end{array} \end{aligned}$

Hence, now the solution of differential equation is,

$\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int(\mathrm{q} \times \mathrm{I} \cdot \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ &\mathrm{y}(\operatorname{Cos} \mathrm{x})=\int(-2 \sin \mathrm{x} \cdot \cos \mathrm{x}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=-\int \sin 2 x \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \cos \mathrm{x}=\frac{\cos 2 x}{2}+c \end{aligned}$

Differential Equations Exercise Revision Exercise Question 42

Answer: $y\: \cos \, x=e^{x}+C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}-y\tan x=e^{x}\sec x$
Solution: $\frac{dy}{dx}-y\tan x=e^{x}\sec x$
Comparing with,
$\begin{aligned} &\frac{\mathrm{dy}}{\text { dx }}+\mathrm{p} \mathrm{y}=\mathrm{q} \ldots \text { (I) we get, }\\ &P=-\tan x \text { and } q=e^{x} \sec x\\ &\text { Now, I.F }=\mathrm{e}^{\int-\tan \mathrm{x} \mathrm{d} \mathrm{x}}\\ &=\mathrm{e}^{-\log |(\operatorname{secx})|}\\ &=\mathrm{e}^{\log \left(\frac{1}{\operatorname{secx}}\right)}\\ &=\mathrm{e}^{\log (\cos x)} \quad\left[\because \mathrm{e}^{\log x}=\mathrm{x}\right]\\ &=\operatorname{Cos} \mathrm{x} \end{aligned}$
So, the solution is,
$\begin{aligned} &\mathrm{y} \mathrm{I} . \mathrm{F}=\int(\mathrm{q} \mathrm{x} \mathrm{I} \cdot \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=\int\left(\cos \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}} \sec \mathrm{x}\right) \mathrm{dx}+\mathrm{C} \\ &=\int \cos \mathrm{x} \cdot \mathrm{e}^{x} \cdot \frac{1}{\cos x} \mathrm{dx}+\mathrm{C} \\ &=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{c} \quad\left[\because \int \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{\mathrm{x}}+\mathrm{C}\right] \\ &\therefore \mathrm{y} \operatorname{Cos} \mathrm{x}=\mathrm{e}^{\mathrm{x}}+\mathrm{C} \end{aligned}$

Differential Equations Exercise Revision Exercise Question 43

Answer: $y\: \cos x=\frac{e^{x}}{2}\left ( \sin x+\cos x \right )+C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\frac{dy}{dx}-y\: \tan \: x=e^{x}$
Solution: $\frac{dy}{dx}-y\: \tan \: x=e^{x}$
Compare with $\frac{dy}{dx}+p\: y=q$ , we get,
$\begin{aligned} &P=-\tan x \text { and } q=e^{x} \\ &\text { Now, } I . F=e^{\int-\tan x d x} \\ &=e^{-\log |(s e c x)|} \\ &=e^{\log \left(\frac{1}{\sec }\right)} \quad\left[\because e^{\log x}=x\right] \\ &\text { I. } F=\frac{1}{\sec x} \\ &=\operatorname{Cos} x \end{aligned}$
Now the solution is,
$\begin{aligned} &\mathrm{y} \mathrm{I} \cdot \mathrm{F}=\int(\mathrm{q} \times \mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=\int\left(\cos \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}}\right)+\mathrm{C} \\ &=\operatorname{Cos} \int \mathrm{e}^{\mathrm{x}} \mathrm{dx}-\int \mathrm{e}^{\mathrm{x}}(-\sin \mathrm{x}) \mathrm{dx}+\mathrm{C} \\ &=\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+\sin \mathrm{x} \int \mathrm{e}^{x}-\int \cos \mathrm{x} \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{C} \\ &=\frac{\mathrm{e}^{\mathrm{x}} \cos x+\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}}{2}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=\frac{\mathrm{e}^{\mathrm{x}}}{2}(\sin \mathrm{x}+\cos \mathrm{x})+\mathrm{C} \end{aligned}$

Differential Equations Exercise Revision Exercise Question 44

Answer: $x\: y=\tan ^{-1}x+C$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$
Solution: $\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$
$\begin{aligned} &x\left(1+x^{2}\right) d y=-\left[\left(1+y\left(1+x^{2}\right)\right] d x\right. \\ &\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{-1-y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}=\frac{-1}{x} \cdot y-\frac{1}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}+\frac{1}{x} \cdot y=-\frac{1}{x\left(1+x^{2}\right)} \end{aligned}$
Comparing with $\frac{dy}{dx}+p\: y=q$ , we get,
$\begin{aligned} &P=\frac{1}{x} \text { and } q=-\frac{1}{x\left(1+x^{2}\right)} \\ &\text { I.F }=e^{\int \frac{1}{x} d x} \\ &=e^{\log x}=x \quad\left[\because \mathrm{e}^{\log x}=x\right] \end{aligned}$
The solution is,
$\begin{aligned} &\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)} \cdot \mathrm{x} \mathrm{dx} \\ &\mathrm{y} \mathrm{x}=-\int \frac{1}{\left(1+\mathrm{x}^{2}\right)} \mathrm{dx} \\ &\mathrm{yx}=-\tan ^{-1} \mathrm{x}+\mathrm{C} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 45

Answer : $y=c e^{-2 \tan ^{-1} x}+\frac{1}{2}$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\left(x^{2}+1\right) d y+(2 y-1) d x=0$
Solution : $\left(x^{2}+1\right) d y+(2 y-1) d x=0$
$\left(x^{2}+1\right) d y=-(2 y-1) d x$
$\begin{aligned} &\frac{d y}{d x}=\frac{(1-2 y)}{\left(1+x^{2}\right)} \\ &\frac{d y}{d x}=\frac{1}{\left(1+x^{2}\right)}-\frac{2 y}{\left(1+x^{2}\right)} \\ &\frac{d y}{d x}+\frac{2 y}{\left(1+x^{2}\right)}=\frac{1}{\left(1+x^{2}\right)} \end{aligned}$
Comparing with $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{p} \mathrm{y}=\mathrm{q}$ , we get,
$\begin{aligned} &\mathrm{P}=\frac{2}{\left(1+\mathrm{x}^{2}\right)} \quad \text { and } \quad \mathrm{q}=\frac{1}{1+\mathrm{x}^{2}} \\ &\text { Now, I.F }=\mathrm{e}^{\int \mathrm{p} \mathrm{dx}} \end{aligned}$
$\begin{aligned} &=e^{\int 2 \frac{d x}{1+x^{2}}} \\ &=e^{2 \tan ^{-1} x} \end{aligned}$
The solution is,
$\begin{aligned} &\mathrm{y} \mathrm{x} \mathrm{I.F}=\int \mathrm{Q} \times \mathrm{I} \cdot \mathrm{F}+\mathrm{C} \\ &\mathrm{y} \mathrm{e}^{2 \tan ^{-1} \mathrm{x}}=\int \frac{\mathrm{e}^{2 \tan ^{-1} \mathrm{x}}}{1+\mathrm{x}^{2}} \mathrm{dx} \end{aligned}$
Put $\tan^{-1}x=t,$ differentiating both side,
$\begin{aligned} &\frac{1}{1+x^{2}} d x=d t \\ &y_{n} e^{2 \tan ^{-1} x}=\int e^{2 t} d t \\ &y e^{2 \tan ^{-1} x}=\frac{1}{2} e^{2 t}+c_{m}\left[\int e^{a t} d x=\frac{1}{a} e^{a x}+c\right] \end{aligned}$
$\begin{aligned} &\text { y } e^{2 \tan ^{-1} x}=\frac{1}{2} e^{2 \tan ^{-1} x}+c \\ &y=c e^{-2 \tan ^{-1} x}+\frac{1}{2} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 46

Answer : $y=\log \frac{(\operatorname{cosec} 2 x-\cot 2 x)}{y^{7}}+c$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\operatorname{ysec}^{2} x+(y+7) \tan x \frac{d y}{d x}=0$
Solution : $\operatorname{ysec}^{2} x+(y+7) \tan x \frac{d y}{d x}=0$
$\begin{aligned} &(y+7) \tan x \frac{d y}{d x}=-\operatorname{ysec}^{2} x \\ &(y+7) d y=\frac{-y \sec ^{2} x}{\tan x} d x \\ &\frac{(y+7)}{y} d y=-\frac{1}{\cos ^{2} x} x \frac{\cos x}{\sin x} d x \end{aligned}$
$\begin{aligned} &\left(1+\frac{7}{y}\right) d y=-\frac{1}{\cos x \sin x} d x \\ &\left(1+\frac{7}{y}\right) d y=-\frac{2}{2 \cos x \sin x} d x \\ &\left(1+\frac{7}{y}\right) d y=-\frac{2}{\sin 2 x} d x \end{aligned}$
Now, Integrating both sides,
$\begin{aligned} &\int 1 d y+\int \frac{7}{y} d y=-\frac{2}{\sin 2 x} d x \\ &y+7 \log |y|=-2 \int \operatorname{cosec} 2 x d x \\ &y+7 \log |y|=-2\left[\frac{1}{2} \log |\operatorname{cosec}(2 x)-\cot (2 x)|\right]+c \end{aligned}$
$\begin{aligned} &\mathrm{y}+\log \mathrm{y}^{7}=\log |\operatorname{cosec}(2 x)-\cot (2 x)|+\mathrm{c} \\ &\mathrm{y}=\log |\operatorname{cosec}(2 x)-\cot (2 x)|-\log \mathrm{y}^{7}+\mathrm{c} \\ &\mathrm{y}=\log \frac{(\operatorname{cosec} 2 \mathrm{x}-\cot 2 \mathrm{x})}{\mathrm{y}^{7}}+\mathrm{c} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 47

Answer : $\mathrm{y}+\mathrm{c}=\frac{\mathrm{a}}{2}[\log |x|-\log |x+2 a|]$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\left(2 a x+x^{2}\right) \frac{d y}{d x}=a^{2}$
Solution : $\left(2 a x+x^{2}\right) \frac{d y}{d x}=a^{2}$
$d y=\left(\frac{a^{2}}{\left(2 a x+x^{2}\right)}\right) d x$
split
$\begin{aligned} &d y=\left(\frac{a^{2}}{x(x+2 a)}\right) d x \\ &d y=\frac{a}{2}\left(\frac{1}{x}-\frac{1}{x+2 a}\right) d x \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int d y=\frac{a}{2}\left[\int \frac{1}{x} d x-\int \frac{1}{x+2 a} d x\right] \\ &y+c=\frac{a}{2}[\log |x|-\log |x+2 a|] \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 48

Answer : $c=4 x^{3} y-2 y^{4}$
Hint: you must know the rules of solving differential equation and integrations.
Given: $\left(x^{3}-2 y^{3}\right) d x+3 x^{2} y d y=0$
Solution : $\left(x^{3}-2 y^{3}\right) d x+3 x^{2} y d y=0$
$\begin{aligned} &3 x^{2} y d y+\left(x^{3}-2 y^{3}\right) d x=0 \\ &3 x^{2} y d y=\left(2 y^{3}-x^{3}\right) d x \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{2 y^{3}-x^{3}}{3 x^{2} y} \\ &\frac{d x}{d y}=\frac{-3 x^{2} y}{x^{3}-2 y^{3}} \\ &=\frac{3}{\frac{-x}{y}+\frac{2 y^{2}}{x^{2}}} \\ &=\frac{3}{2\left(\frac{y}{x}\right)^{2}-\frac{x}{y}} \end{aligned}$
Put $\frac{y}{x}=y$ , differentiate both sides,
$y=vx$
$\begin{aligned} &\frac{d y}{d x}=v+\frac{x d v}{d x} \\ &v+\frac{x d v}{d x}=\frac{3}{2 v^{2}-\frac{1}{v}} \\ &\frac{x d v}{d x}=\frac{3 v-2 v^{4}+v}{2 v^{3}-1} \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{2 \mathrm{v}^{3}-1}{3 \mathrm{v}-2 \mathrm{v}^{4}+\mathrm{v}} \mathrm{d} \mathrm{V} \\ &\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{2 \mathrm{v}^{3}-1}{4 \mathrm{v}-2 \mathrm{v}^{4}} \mathrm{~d} \mathrm{~V} \end{aligned}$
put, $4 v-2 v^{4}=t$
$\begin{aligned} &\left(4-8 v^{3}\right) d v=d t \\ &-4\left(2 v^{3}-1\right) d v=d t \\ &\int \frac{d x}{x}=\frac{-1}{4} \int \frac{d t}{t} \\ &\log x+\log c=\frac{-1}{4} \log t \end{aligned}$
$\begin{aligned} &\log (c x)=\log t^{\frac{-1}{4}} \\ &t^{\frac{-1}{4}}=c x \end{aligned}$
Put value of t
$\left(4 v-2 v^{4}\right)^{\frac{-1}{4}}=c x$
put value of $v=\frac{y}{x}$
$\begin{aligned} &\Rightarrow\left(\frac{4 y}{x}-\frac{2 y^{4}}{x^{4}}\right)^{\frac{-1}{4}}=\mathrm{cx} \\ &\Rightarrow\left(\frac{4 x^{3} y-2 y^{4}}{x^{4}}\right)^{\frac{-1}{4}}=c x \\ &\Rightarrow \frac{x}{\left(4 x^{3} y-2 y^{4}\right)^{\frac{1}{4}}}=c x \quad\left[\because\left(x^{4}\right)^{\frac{-1}{4}}=x^{-1}\right] \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{\left(4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4}\right)^{\frac{1}{4}}}=\mathrm{c} \\ &\Rightarrow \frac{1}{\mathrm{c}}=\left(4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4}\right)^{\frac{1}{4}} \\ &\Rightarrow \sqrt[4]{\mathrm{c}}=4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4} \\ &\Rightarrow \quad \mathrm{c}=4 \mathrm{x}^{3} \mathrm{y}-2 \mathrm{y}^{4} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 49

Answer : $c=x e^{\tan ^{-1} \frac{y}{x}}$
Hint: you must know the rules of solving differential equation and integrations.
Given: $x^{2} d y+\left(x^{2}-x y+y^{2}\right) d x=0$
Solution : $x^{2} d y+\left(x^{2}-x y+y^{2}\right) d x=0$
$\begin{aligned} &\Rightarrow x^{2} d y=-\left(x y-x^{2}-y^{2}\right) d x \\ &\Rightarrow \frac{d y}{d x}=\frac{x y-x^{2}-y^{2}}{x^{2}} \end{aligned}$
putting $y=vx$ and differentiate,
$\begin{aligned} &\frac{d y}{d x}=v+x \frac{d v}{d x} \\ &\left.v+x \frac{d v}{d x}=\frac{x y-x^{2}-y^{2}}{x^{2}} \quad \text { [put value of } \frac{d y}{d x}\right] \\ &v+x \frac{d v}{d x}=v-1-v^{2} \end{aligned}$
$\begin{aligned} &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-1-\mathrm{v}^{2}-\mathrm{v} \\ &\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-1-\mathrm{v}^{2} \\ &\frac{\mathrm{dv}}{1+\mathrm{v}^{2}}=-\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{d v}{1+v^{2}} d v=-\int \frac{1}{x} d x \\ &\tan ^{-1} v=-\log |x|+\log c\left[\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x\right] \\ &\tan ^{-1} v=\log \frac{c}{x} \end{aligned}$
$\begin{aligned} &\mathrm{e}^{\tan ^{-1} v}=\frac{c}{\mathrm{x}}\left[\log \mathrm{x}=\mathrm{a}, \mathrm{x}=\mathrm{e}^{\mathrm{a}}\right]\\ &\left.c=x e^{\tan ^{-1} \frac{y}{x}} \text { [put value of } v=\frac{x}{y}\right] \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 50

Answer : $x=c^{1}(y-b)(b x+1)$
Hint: you must know the rules of solving differential equation and integrations.
Given: $(y-x) \frac{d y}{d x}=b\left(1+x^{2} \frac{d y}{d x}\right)$
Solution : $(y-x) \frac{d y}{d x}=b\left(1+x^{2} \frac{d y}{d x}\right)$
$\begin{aligned} &y-b=\left(b x^{2}+x\right) \frac{d y}{d x} \\ &\left(\frac{1}{y-b}\right) d y=\left(\frac{1}{b x^{2}+x}\right) d x \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{1}{y-b} d y=\int \frac{1}{b x^{2}+x} d x \\ &\int \frac{1}{y-b} d y=\frac{1}{b} \int \frac{1}{x^{2}+\frac{x}{b}} d x \\ &\int \frac{1}{y-b} d y=\frac{1}{b} \int\left(\frac{1}{x^{2}+\frac{x}{b}+\frac{1}{4 b^{2}}-\frac{1}{4 b^{2}}}\right) d x \end{aligned}$
$\int\left(\frac{1}{y-b}\right) d y=\frac{1}{b} \int\left(\frac{1}{\left(x+\frac{1}{2 b}\right)^{2}-\left(\frac{1}{2 b}\right)^{2}}\right) d x$
$\begin{aligned} &\Rightarrow \log |y-b|=\frac{1}{2 \operatorname{x} \frac{1}{2 \mathrm{~b}} \times \mathrm{b}} \log \left|\frac{x+\frac{1}{2 b}-\frac{1}{2 b}}{x+\frac{1}{2 b}+\frac{1}{2 b}}\right|+\log c\left[\because \int \frac{1}{\mathrm{x}} \mathrm{dx}=\log |\mathrm{x}| \text { and } \int \frac{1}{(\mathrm{x}+\mathrm{a})^{2}-\mathrm{a}^{2}}=\log \left|\frac{x+a-a}{x+a+a}\right|\right] \\ &\Rightarrow \log |y-b|=\log \left|\frac{b x}{b x+1}\right|+\log \mathrm{c} \end{aligned}$
$\begin{aligned} & \Rightarrow \quad y-b=\frac{c b x}{b x+1} \\ & \Rightarrow \quad c b x=(y-b)(b x+1) \\ \therefore x &=c^{1}(y-b)(b x+1) \quad\left[\text { where } \frac{1}{c b}=c^{1}\right] \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 51

Answer : $y=\frac{1}{13}(2 \sin 3 x-3 \cos 3 x)+C e^{-2 x}$
Hint : You must know the rules of solving differential equation and integration.
Given : $\frac{d y}{d x}+2 y=\sin 3 x$
Solution : $\frac{d y}{d x}+2 y=\sin 3 x$
Compare with,
$\frac{d y}{d x}+P y=Q$
where, $P=2$ and $Q= \sin 3x$
Therefore,
$\begin{aligned} \text { Integrating factor }=\text { I.F } &=e^{\int p d x} \\ &=e^{\int 2 d x} \\ &=e^{2 x} \end{aligned}$
The solution is,
$\begin{aligned} &y \times I . F=\int(Q \times I . F) d x+C \\ &y e^{2 x}=\int e^{2 x} \times \sin 3 x d x+C \\ &y e^{2 x}=I+C \end{aligned}$ ......(i)
Where, $I=\int e^{2 x} \sin 3 x d x$ ......(ii)
Apply integrating by parts,
$\begin{aligned} I &=e^{2 x} \int \sin 3 x d x-\int\left[\frac{d e^{2 x}}{d x} \int \sin 3 x d x\right] d x \\ I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3} \int e^{2 x} \cos 3 x d x \\ I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3}\left[e^{2 x} \int \cos 3 x d x-\int\left\{\frac{d e^{2 x}}{d x} \int \cos 3 x d x\right\} d x\right] \end{aligned}$
$\begin{aligned} I &=\frac{-e^{2 x} \cos 3 x}{3}+\frac{2}{3}\left[\frac{e^{2 x} \sin 3 x}{3}-\frac{2}{3} \int e^{2 x} \sin 3 x d x\right] \\ I &=\left[\frac{2}{9} e^{2 x} \sin 3 x-\frac{e^{2 x} \cos 3 x}{3}\right]-\frac{4}{9} I \end{aligned}$
$\begin{aligned} I+\frac{4}{9} I &=e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ \frac{13}{9} I &=e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ I &=\frac{9}{13} e^{2 x}\left(\frac{2}{9} \sin 3 x-\frac{\cos 3 x}{3}\right) \\ I &=\frac{1}{13} e^{2 x}(2 \sin 3 x-3 \cos 3 x)+C \end{aligned}$ .....(iii)
From (i) and (iii), we get,
$\begin{aligned} y e^{2 x} &=\frac{e^{2 x}}{13}(2 \sin 3 x-3 \cos 3 x)+C \\ y \quad &=\frac{1}{13}(2 \sin 3 x-3 \cos 3 x)+C e^{-2 x} \end{aligned}$ Is required solution.

Differential Equation Exercise Revision Exercise (RE) Question 52

Answer : $y=4(x-1)+c e^{-x}$
Hints : You must know the rules of solving differential equation and integration.
Given : $\frac{dy}{dx}+y=4x$
Solution : $\frac{dy}{dx}+y=4x$ .....(i)
Compare with, $\frac{dy}{dx}+py=Q$
Where, $P=1, Q=4x$
Therefore,
$\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\qquad=e^{\int d x} \\ &\; \; \; \; \; \; \; =e^{x} \end{aligned}$
Hence, the solution is ,
$\begin{aligned} y \times I . F &=\int(I . F \times Q) d x+c \\ y \quad e^{x} &=\int e^{x} 4 x d x+c \end{aligned}$
Integrating by parts,
$\begin{aligned} &y e^{x}=4 x \int e^{x} d x-4 \int\left[\frac{d}{d x}(x) \int e^{x} d x\right]+c \\ &y e^{x}=4 x e^{x}-4 \int e^{x} d x+c \\ &y e^{x}=4 x e^{x}-4 e^{x}+c \end{aligned}$
$\begin{aligned} y e^{x} &=4(x-1) e^{x}+c \\ y &=4(x-1)+c e^{-x} \end{aligned}$ is required solution

Differential Equation Exercise Revision Exercise (RE) Question 53

Answer : $\frac{1}{41}(4 \sin 4 x+5 \cos 4 x)+c e^{-5 x}$
Hint : You must know the rules of solving differential equation and integration.
Given : $\frac{dy}{dx}+5y=\cos 4x$
Solution : $\frac{dy}{dx}+5y=\cos 4x$
Compare with,
$\frac{dy}{dx}+Py=Q$
When $P=5, Q= \cos 4x$
Therefore,
$\begin{aligned} I . F &=e^{\int P d x} \\ &=e^{\int 5 d x} \\ &=e^{5 x} \end{aligned}$
The solution is ,
$\begin{aligned} y \times I . F &=\int(I . F \times Q) d x+c \\ y e^{5 x} &=\int e^{5 x} \times \cos 4 x d x+c \\ y e^{x} &=I+c \end{aligned}$ ....(i)
Where,
$\begin{aligned} I &=\int e^{5 x} \cos 4 x d x \\ I &=e^{5 x} \int \cos 4 x d x-\int\left[\frac{d e^{5 x}}{d x} \int \cos 4 x d x\right] d x \\ I &=e^{5 x} \frac{\sin 4 x}{4}-\frac{5}{4} \int e^{5 x} \sin 4 x d x \end{aligned}$
$\begin{aligned} &I=\frac{e^{5 x} \sin 4 x}{4}-\frac{5}{4}\left[e^{5 x} \int \sin 4 x d x-\int\left\{\frac{d e^{5 x}}{d x} \int \sin 4 x d x\right\} d x\right] \\ &I=\frac{e^{5 x} \sin 4 x}{4}-\frac{5}{4}\left[\frac{-e^{5 x} \cos 4 x}{4}+\frac{5}{4} \int e^{5 x} \cos 4 x d x\right] \\ &I=\frac{e^{5 x} \sin 4 x}{4}+\frac{5}{16} e^{5 x} \cos 4 x-\frac{25}{16} \int e^{5 x} \cos 4 x d x \end{aligned}$
$\begin{gathered} I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x)-\frac{25}{16} I \\ \frac{25}{16} I+I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x) \\ \frac{41}{16} I=\frac{e^{5 x}}{16}(4 \sin 4 x+5 \cos 4 x) \\ I=\frac{e^{5 x}}{41}(4 \sin 4 x+5 \cos 4 x) \end{gathered}$
Therefore, required solution is
$\begin{aligned} y e^{5 x} &=\frac{e^{5 x}}{41}(4 \sin 4 x+5 \cos 4 x)+c \\ y &=\frac{1}{41}(4 \sin 4 x+5 \cos 4 x)+c e^{-5 x} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 54

Answer : $\tan \frac{y}{x}=-\log |x|+c$
Hint : You must know the rules of solving differential equation and integration
Given : $x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y$
Solution : $x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y$
$\begin{aligned} &\frac{d y}{d x}+\cos ^{2}\left(\frac{y}{x}\right)=\frac{y}{x} \\ &\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right) \end{aligned}$
Putting $y=vx$ and differentiate
Therefore, $\frac{d y}{d x}=v+x \frac{d v}{d x}$
Therefore,
$\begin{aligned} &v+x \frac{d v}{d x}=v-\cos ^{2}(v) \\ &\frac{1}{\cos ^{2} v} d v=-\frac{d x}{x} \\ &\sec ^{2} v d v=-\frac{1}{x} d x \end{aligned}$
Integrating both sides,
$\begin{aligned} \int \sec ^{2} d v &=-\int \frac{1}{x} d x \\ \tan v &=-\log |x|+c \\ \tan \frac{y}{x} &=-\log |x|+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 55

Answer : $y e^{\tan x}=e^{\tan x}(\tan x-1)+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\cos ^{2} x \frac{d y}{d x}+y=\tan x$
Solution : $\cos ^{2} x \frac{d y}{d x}+y=\tan x$
$\begin{aligned} \frac{d y}{d x}+\frac{y}{\cos ^{2} x} &=\frac{\tan x}{\cos ^{2} x} \\ \frac{d y}{d x}+\left(\sec ^{2} x\right) y &=(\tan x) \sec ^{2} x \end{aligned}$
Comparing with ,
$\frac{dy}{dx}+Py=Q$
Where, $\mathrm{P}=\sec ^{2} x, \mathrm{Q}=(\tan x)\left(\sec ^{2} x\right)$
Now,
$\begin{aligned} \text { I.F } &=e^{\int \sec ^{2} x d x} \\ &=e^{\tan x} \end{aligned}$
So, the solution is
$\begin{aligned} y \times I . F &=\int(I . F \times Q) d x+c \\ y \times e^{\tan x} &=\int e^{\tan x} \times(\tan x)\left(\sec ^{2} x\right) d x+c \\ y e^{\tan x} &=I+C \end{aligned}$
Now,
$I=\int \tan x\left(\sec ^{2} x\right) \times e^{\tan x} d x$
Integrating by parts,
$\begin{aligned} &\text { Put } t=\tan x \\ &d t=\sec ^{2} x d x \end{aligned}$
Therefore,
$\begin{aligned} I &=\int\left(t \times e^{t}\right) d t \\ I &=t \times \int e^{t} d t-\int\left(\frac{d t}{d t} \times \int e^{t} d t\right) d t \\ &=t e^{t}-\int e^{t} d t \\ &=t e^{t}-e^{t} \end{aligned}$
Therefore,
$\begin{aligned} I &=\tan x e^{\tan x}-e^{\tan x} \\ &=e^{\tan x}(\tan x-1) \end{aligned}$
Hence, the required solution is
$y e^{\tan x}=e^{\tan x}(\tan x-1)+c$

Differential Equation Exercise Revision Exercise (RE) Question 56

Answer : $x y \sec x=\tan x+c$
Hint : You must know the rules of solving differential equation and integration
Given : $x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1$
Solution :
$\begin{aligned} x \cos x \frac{d y}{d x}+y(x \sin x+\cos x) &=1 \\ \frac{d y}{d x}+y\left(\frac{x \sin x}{x \cos x}+\frac{\cos x}{x \cos x}\right) &=\frac{1}{x \cos x} \\ \frac{d y}{d x}+\left(\tan x+\frac{1}{x}\right) y &=\frac{1}{x \cos x} \end{aligned}$
Comparing with, $\frac{d y}{d x}+P y=Q$
Where, $P=\tan x+\frac{1}{x}, Q=\frac{1}{x \cos x}$
Therefore,
$\begin{aligned} I . F &=e^{\int\left(\tan x+\frac{1}{x}\right) d x} \\ &=e^{\log |\sec x|+\log |x|} \\ &=e^{\log |x \sec x|} \\ &=|x \sec x| \end{aligned}$ $\left[e^{\log x}=x\right]$
So the solution is,
$\begin{aligned} y \times I . F &=\int(I . F) \times(Q) d x+C \\ x y \sec x &=\int x \sec x \times \frac{1}{x \cos x} d x+c \\ &=\int \sec ^{2} x+c \\ x y \sec x &=\tan x+c \end{aligned}$ $\left[\int \sec ^{2} x d x=\tan x\right]$

Differential Equation Exercise Revision Exercise (RE) Question 57

Answer : $x e^{\tan ^{-1} y}=\tan ^{-1} y+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\left(1+y^{2}\right)+\left(x-e^{-\tan ^{-1} y}\right) \frac{d y}{d x}=0$
Solution : $\left(1+y^{2}\right)+\left(x-e^{-\tan ^{-1} y}\right) \frac{d y}{d x}=0$
$\begin{aligned} &\frac{d x}{d y}=\frac{e^{-\tan ^{-1} y}-x}{1+y^{2}} \\ &\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{-\tan ^{-1} y}}{1+y^{2}} \end{aligned}$
Comparing with,
$\frac{d x}{d y}+P x=Q, we get$
$P=\frac{1}{1+y^{2}} \quad, Q=\frac{e^{-\tan ^{-1} y}}{1+y^{2}}$
Now, $\text { I.F }=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$
So, the solution is
$\begin{aligned} x \times I . F &=\int I . F \times Q d y+c \\ x \times e^{\tan ^{-1} y} &=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \times e^{-\tan ^{-1} y} d y+c \\ x \times e^{\tan ^{-1} y} &=\int \frac{1}{1+y^{2}} d y+c \\ x e^{\tan ^{-1} y} &=\tan ^{-1} y+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 58

Answer : $x=-\left(1+\frac{1}{y}\right)+c e^{\frac{1}{y}}$
Hint : you must know the rules of solving differential equation and integration
Given : $y^{2}+\left(x+\frac{1}{y}\right) \frac{d y}{d x}=0$
Solution : $y^{2}+\left(x+\frac{1}{y}\right) \frac{d y}{d x}=0$
$\begin{aligned} &\frac{d y}{d x}=\frac{-y^{3}}{x y+1} \\ &\frac{d x}{d y}=\frac{x y+1}{-y^{3}} \\ &\frac{d x}{d y}=\frac{-x}{y^{2}}-\frac{1}{y^{3}} \\ &\frac{d x}{d y}+\frac{x}{y^{2}}=\frac{-1}{y^{3}} \end{aligned}$
Comparing with, $\frac{d x}{d y}+P x=Q$
Where , $P=\frac{1}{y^{2}}, Q=\frac{-1}{y^{3}}$
Now,
$\text { I. } F=e^{\int \frac{1}{y^{2}} d y}=e^{\frac{-1}{y}}$
therefore the solution is
$\begin{aligned} x \times I . F &=\int \text { I.F } \times Q d y+c \\ x e^{\frac{-1}{y}} &=\int-e^{\frac{-1}{y}} \frac{1}{y^{3}} d y+c \\ x e^{\frac{-1}{y}} &=I+c \end{aligned}$
Putting $t=\frac{1}{y},$ and differentiating both sides
$d t=\frac{-1}{y^{2}} d y$
Applying integration both side,
$\begin{aligned} I &=\int\left(t \times e^{-t}\right) d t \\ I &=t \times \int e^{-t} d t-\int\left(\frac{d t}{d t} \times \int e^{-t} d t\right) d t \\ &=-t e^{-t}+\int e^{-t} d t \\ &=-t e^{-t}-e^{-t} \end{aligned}$
Therefore,
$\begin{aligned} &I=\frac{-1}{y} e^{\frac{-1}{y}}-e^{\frac{-1}{y}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[t=\frac{1}{y}\right] \\ &I=-e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right) \end{aligned}$
Hence, required solution is ,
$\begin{aligned} &x e^{\frac{-1}{y}}=-e^{\frac{-1}{y}}\left(1+\frac{1}{y}\right)+c \\ &x=-\left(1+\frac{1}{y}\right)+c e^{\frac{1}{y}} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 59

Answer : $y=\cos x-2 \cos ^{2} x$
Hint : you must know the rules of solving differential equation and integration
Given : $2 \cos x \frac{d y}{d x}+4 y \sin x=\sin 2 x$ , $y=0, x=\frac{\Pi}{3}$
Solution : $2 \cos x \frac{d y}{d x}+4 y \sin x=\sin 2 x$
$\begin{aligned} &\frac{d y}{d x}+4 y \frac{\sin x}{2 \cos x} \quad=\frac{2 \sin x \cos x}{2 \cos x} \\ &\frac{d y}{d x}+2 y \tan x \quad=\sin x \end{aligned}$
Comparing with, $\frac{d y}{d x}+P y=Q$ , we get
Where, $P=2\tan x, Q = \sin x$
Now,
$\begin{aligned} \text { I.F } &=e^{2 \int \tan x d x} \\ &=e^{2 \log |\sec x|} \\ &=\sec ^{2} x \end{aligned}$
So, the solution is,
$\begin{aligned} &y \times \text { I.F }=\int(Q \times I . F) d x+c \\ &y \sec ^{2} x=\int \sin x \sec ^{2} x d x+c \\ &y \sec ^{2} x=\int \tan x \sec x d x+c \end{aligned}$
$\begin{aligned} &y \sec ^{2} x=\sec x+c \\ &y=\cos x+C \cos ^{2} x \\ &\text { now } x=\frac{\pi}{3} \quad, y=0 \end{aligned}$
Therefore,
$\begin{aligned} &0=\cos \frac{\pi}{3}+C \cos ^{2} \frac{\pi}{3} \\ &0=\frac{1}{2}+C\left(\frac{1}{4}\right) \end{aligned}$
$C=-2$
Putiing value of C,
$y=\cos x-2 \cos ^{2} x$

Differential Equation Exercise Revision Exercise (RE) Question 60

Answer : $x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y$
Solution : $\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y$
$\begin{aligned} \frac{d x}{d y} &=\frac{\tan ^{-1} y-x}{1+y^{2}} \\ \frac{d x}{d y}+\frac{x}{1+y^{2}} &=\frac{\tan ^{-1} y}{1+y^{2}} \end{aligned}$
Comparing with, $\frac{dx}{dy}+Px=Q,$
$\begin{aligned} &P=\frac{1}{1+y^{2}}, Q=\frac{\tan ^{-1} y}{1+y^{2}} \\ &\text { Now } \quad \text { I. } F=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y} \end{aligned}$
So, the solution is ,
$\begin{aligned} x \times I . F &=\int(I . F \times Q) d y+c \\ x e^{\tan ^{-1} y} &=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y+c \\ x e^{\tan ^{-1} y} &=I+C \end{aligned}$
Now, $I=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y$
$\begin{aligned} &\text { Put }_{\mu} \text { tan }^{-1} y=t \text { and differentiate, }\\ &\begin{gathered} \quad \frac{1}{1+y^{2}} d y=d t \\ I=\int e^{t} t d t \end{gathered} \end{aligned}$
Integrating by parts,
$\begin{aligned} &I=t \times \int e^{t} d t-\int\left\{\frac{d t}{d t} \int e^{t} d t\right\} d t \\ &I=t e^{t}-\int e^{t} d t \\ &I=t e^{t}-e^{t} \end{aligned}$
$\begin{aligned} &I=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y} \\ &I=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right) \end{aligned}$
By putting value of I
We get the required solution,
$x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c$

Differential Equation Exercise Revision Exercise (RE) Question 61

Answer : $y \sec x=\frac{x^{n+1}}{n+1}+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}+y \tan x=x^{n} \cos x$
Solution : $\frac{d y}{d x}+y \tan x=x^{n} \cos x$
Comparing with, $\frac{d y}{d x}+Py=Q$
$P=\tan x \quad, Q=x^{n} \cos x$
Now,
$\begin{aligned} \text { I.F } &=e^{\int \tan x d x} \\ &=e^{\log (\sec x)} \\ &=\sec x \end{aligned}$
So the solution is ,
$\begin{aligned} &y \times \text { I.F }=\int(Q \times I . F) d x+c \\ &y \sec x=\int x^{n} \cos x \sec x d x+c \\ &y \sec x=\int x^{n} d x+c \\ &y \sec x=\frac{x^{n+1}}{n+1}+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 62

Answer : $x^{2}+y^{2}+2 x-4 y+c=0$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=\frac{x+1}{2-y}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad, y \neq 2$
Solution : $\frac{d y}{d x}=\frac{x+1}{2-y}$
$(2-y) d y=(x+1) d x$
Integrating both sides,
$\begin{array}{r} \int(2-y) d y=\int(x+1) d x \\ 2 y-\frac{y^{2}}{2}=\frac{x^{2}}{2}+x+c_{1} \\ \frac{x^{2}}{2}+x+c_{1}-2 y+\frac{y^{2}}{2}=0 \end{array}$
$\begin{array}{ll} x^{2}+2 x+y^{2}-4 y+2 c_{1} & =0 \\ x^{2}+y^{2}+2 x-4 y+c & =0\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[2 c_{1}=c\right] \end{array}$

Differential Equation Exercise Revision Exercise (RE) Question 63

Answer : $y=\frac{1}{2 x^{2}+1}$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=-4 x y^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad y=1, x=0$
Solution : $\frac{d y}{d x}=-4 x y^{2}$
$\frac{1}{y^{2}}dy=-4x \; dx$
Integrating both sides ,
$\begin{gathered} \int \frac{1}{y^{2}} d y=-4 \int x d x \\ \Rightarrow \frac{-1}{y}=-2 x^{2}+c \end{gathered}$
Now, $x=0, y=1$
Therefore, $-1= 0+c$
$c= -1$
Put value of c,
$\begin{aligned} &\frac{-1}{y}=-2 x^{2}-1 \\ &\frac{1}{y}=2 x^{2}+1 \\ &y=\frac{1}{2 x^{2}+1} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 64 (i)

Answer : $y=2 \tan \frac{x}{2}-x+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$
Solution : $\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}$
$\begin{aligned} &\frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \\ &\frac{d y}{d x}=2 \tan ^{2} \frac{x}{2} \end{aligned}$
$\begin{aligned} &d y=\left(\tan ^{2} \frac{x}{2}\right) d x \\ &d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x \end{aligned}$
Integrating both sides,
$\begin{gathered} \int d y=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x \\ \int d y=\int \sec ^{2} \frac{x}{2}-\int 1 d x \\ y=2 \tan \frac{x}{2}-x+c \end{gathered}$

Differential Equation Exercise Revision Exercise (RE) Question 64 (ii)

Answer : $y=2 \sin (x+c)$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=\sqrt{4-y^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad,-2<y<2$
Solution : $\frac{d y}{d x}=\sqrt{4-y^{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad,-2<y<2$
$\frac{1}{\sqrt{4-y^{2}}}dy=dx$
Integrating both sides,
$\begin{aligned} \int \frac{1}{\sqrt{4-y^{2}}} d y &=\int d x \\ \sin ^{-1} \frac{y}{2} &=x+c \\ \frac{y}{2} &=\sin (x+c) \\ y &=2 \sin (x+c) \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 64 (iii)

Answer : $\tan ^{-1} y=x+\frac{x^{3}}{3}+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$
Solution : $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$
$\frac{1}{1+y^{2}} d y=\left(1+x^{2}\right) d x$
Integrating both sides,
$\begin{gathered} \int \frac{1}{1+y^{2}} d y=\int\left(1+x^{2}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{1+y^{2}} d y=\tan ^{-1} y+c\right] \\ \end{gathered}$
$\tan ^{-1} y=x+\frac{x^{3}}{3}+c$

Differential Equation Exercise Revision Exercise (RE) Question 64 (iv)

Answer : $y=e^{cx}$
Hint : You must know the rules of solving differential equation and integration
Given : $y \log y d x-x d y=0$
Solution : $y \log y d x-x d y=0$
$y \log y d x=x\; dy$
$\begin{aligned} \frac{1}{x} d x &=\frac{1}{y \log y} d y \\ \frac{1}{y \log y} d y &=\frac{1}{x} d x \end{aligned}$
Integrating both sides,
$\int \frac{1}{y \log y} d y=\int \frac{d x}{x}$
Put $log\; y =t$ and differentiating,
$\frac{1}{y}dy=dt$
Therefore,
$\begin{aligned} &\int \frac{1}{t} d t=\int \frac{1}{x} d x \\ &\log |t|=\log x+\log c \\ &\log (\log y)=\log x+\log c \end{aligned}$
$\begin{aligned} &\log (\log y)=\log c x \\ &\log y=c x \\ &y=e^{c x} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 64 (v)

Answer : $y=x \sin ^{-1} x+\sqrt{1-x^{2}}+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=\sin ^{-1} x$
Solution : $\frac{d y}{d x}=\sin ^{-1} x$
$\frac{d y}{d x}=\sin ^{-1} x\; dx$
Integrating both sides
$\int d y=\int 1 \times \sin ^{-1} x d x$
Integrating by parts,
Put $\begin{aligned} \int d y &=\sin ^{-1} x \int 1 d x-\int\left[\frac{d}{d x}\left(\sin ^{-1} x\right) \int 1 d x\right] d x \\ y &=x \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x \\ t^{2} &=1-x^{2}, \text { and differentiate we get } \end{aligned}$
$\begin{gathered} 2 t d t=-2 x d x \\ \Rightarrow-t d t=x d x \end{gathered}$
Therefore,
$\begin{aligned} &y=x \sin ^{-1} x+\int d t \\ &y=x \sin ^{-1} x+t+c \\ &y=x \sin ^{-1} x+\sqrt{1-x^{2}}+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 64 (vi)

Answer : $y=1+C e^{-x}$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}+y=1$
Solution : $\frac{d y}{d x}+y=1$
$\begin{aligned} &\frac{d y}{d x}=1-y \\ &\frac{1}{1-y} d y=d x \end{aligned}$
Integrating both sides
$\begin{aligned} &\int \frac{1}{(1-y)} d y=\int d x \\ &-\int \frac{1}{y-1} d y=\int d x \\ &\int \frac{1}{y-1} d y=-\int d x \end{aligned}$
$\begin{aligned} &\log |y-1|=-x+\log c \\ &\log |y-1|-\log c=-x \\ &\log \left|\frac{y-1}{c}\right|=-x \end{aligned}$
$\begin{aligned} &\frac{y-1}{c}=e^{-x} \\ &y=1+C e^{-x} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 65 (i)

Answer : $y=\frac{1}{2} \log \left(\frac{x^{2}-1}{x^{2}}\right)-\frac{1}{2} \log \left(\frac{3}{4}\right)$
Hint : You must know the rules of solving differential equation and integration
Given : $x\left(x^{2}-1\right) \frac{d y}{d x}=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad, y=0 \text { where } x=2$
Solution : $x\left(x^{2}-1\right) \frac{d y}{d x}=1$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{x\left(x^{2}-1\right)} \\ &d y=\left\{\frac{1}{x\left(x^{2}-1\right)}\right\} d x \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int d y=\int\left\{\frac{1}{x\left(x^{2}-1\right)}\right\} d x \\ &y=\int \frac{1}{x(x+1)(x-1)} d x+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right] \end{aligned}$
Let,
$\begin{gathered} \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1} \\ 1=A(x+1)(x-1)+B x(x-1)+C x(x+1) \\ 1=A\left(x^{2}-1\right)+B\left(x^{2}-x\right)+C\left(x^{2}+x\right) \\ 1=x^{2}(A+B+C)+x(-B+C)-A \end{gathered}$
Comparing both sides,
$\begin{array}{r} -A=1 \\ -B+C=0 \\ A+B+C=0 \end{array}$
Therefore, $A=-1, B=\frac{1}{2}, C =\frac{1}{2}$
Therefore,
$\frac{1}{x(x+1)(x-1)}=\frac{-1}{x}+\frac{1}{2(x+1)}+\frac{1}{2(x-1)}$
Now,
$\begin{aligned} &y=\int\left(\frac{-1}{x}+\frac{1}{2(x+1)}+\frac{1}{2(x-1)}\right) d x+c \\ &y=-\log |x|+\frac{1}{2} \log |x+1|+\frac{1}{2} \log |x-1|+c \\ &y=\frac{1}{2} \log |x+1|+\frac{1}{2} \log |x-1|-\log |x|+c \end{aligned}$
Given, $y(2)=0$
Therefore, $y=0,x=2$
$\begin{aligned} &0=\frac{1}{2} \log |2+1|+\frac{1}{2} \log |2-1|-\log |2|+c \\ &c=\log |2|-\frac{1}{2} \log (3) \quad[\log 1=0] \end{aligned}$
Therefore the solution is ,
$\begin{aligned} &y=\frac{1}{2} \log |x-1|+\frac{1}{2} \log |x+1|-\log |x|+\log (2)-\frac{1}{2} \log (3) \\ &2 y=\log |x-1|+\log |x+1|-2 \log |x|+2 \log (2)-\log 3 \\ &2 y=\log |x-1|+\log |x+1|-\log x^{2}+\log \left(2^{2}\right)-\log 3 \end{aligned}$
$\begin{aligned} &2 y=\frac{\log (x-1)(x+1)}{x^{2}}-[\log (3)-\log (4)] \\ &y=\frac{1}{2} \log \frac{x^{2}-1}{x^{2}}-\frac{1}{2} \log \left(\frac{3}{4}\right) \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 65 (ii)

Answer : $\cos \left(\frac{y-1}{x}\right)=a$
Hint : You must know the rules of solving differential equation and integration
Given : $\cos \left(\frac{d y}{d x}\right)=a$
Solution : $\cos \left(\frac{d y}{d x}\right)=a$
$\begin{aligned} &\frac{d y}{d x}=\cos ^{-1} a \\ &d y=\cos ^{-1} a d x \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int d y=\int \cos ^{-1} a d x \\ &\int d y=\cos ^{-1} a \int d x \\ &y=x \cos ^{-1} a+c \end{aligned}$
Now, $x=0,y=1$
Therefore, $1=0+C$
$C=1$
Hence,
$\begin{aligned} &y=x \cos ^{-1} a+1 \\ &\cos \left(\frac{y-1}{x}\right)=a \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 65 (iii)

Answer : $y=\sec x$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}=y \text { tan } x \quad, y=1 \text { where }, x=0$
Solution : $\frac{d y}{d x}=y \text { tan } x$
$\frac{1}{y} d y=\tan x d x$
Integrating both sides,
$\begin{aligned} &\int \frac{1}{y} d y=\int \tan x d x \\ &\Rightarrow \log y=\log |\sec x|+C \end{aligned}$
Now, $x=0,y=1$
Therefore,
$\begin{gathered} \log 1=\log 1+C \\ C=0 \end{gathered}$
Put value of c ,
$\begin{aligned} \log y &=\log |\sec x|+0 \\ \log y &=\log |\sec x| \\ y &=\sec x \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (i)

Answer : $\log \left|x^{2}+x y+y^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+c$
Hint : You must know the rules of solving differential equation and integration
Given : $(x-y) \frac{d y}{d x}=x+2 y$
Solution :
STEP : 1
$\frac{dy}{dx}=\frac{x+2y}{x-y}$
STEP : 2
$\begin{aligned} &\text { Put } f(x)=\frac{d y}{d x} \quad, \text { find } f(\lambda x, \lambda y) \\ &\frac{d y}{d x}=\frac{x+2 y}{x-y} \\ &\text { Put } f(x, y)=\frac{x+2 y}{x-y} \\ &\text { Find } f(\lambda x, \lambda y) \end{aligned}$
$\begin{aligned} f(\lambda x, \lambda y) &=\frac{\lambda x+2(\lambda y)}{\lambda x-\lambda y} \\ &=\frac{\lambda(x+2 y)}{\lambda(x-y)} \\ &=\frac{x+2 y}{x * y} \end{aligned}$ .....(i)
$=f(x,y)$
STEP : 3
$\text {Put}\; y=vx$
$\begin{aligned} &\frac{d y}{d x}=\frac{d(v x)}{d x} \\ &\text { So } \frac{d y}{d x}=\frac{d v}{d x} \times x+v \frac{d x}{d x} \\ &\frac{d y}{d x}=x \frac{d v}{d x}+v \\ &\text { Put } \frac{d y}{d x} \text { and } \frac{y}{x} \text { in }(i) \end{aligned}$
$\begin{aligned} &\frac{d v}{d x} x+v=\frac{x+2 v x}{x-v x} \\ &\frac{d v}{d x} x+v=\frac{x(1+2 v)}{x(1-v)} \\ &\frac{d v}{d x} x=\frac{(1+2 v-v(1-v))}{1-v} \end{aligned}$
$\begin{gathered} \frac{d v}{d x} x=\frac{1+2 v-v+v^{2}}{1-v} \\ \frac{d v}{d x} x=\frac{v^{2}+v+1}{1-v} \\ \frac{d v}{d x} x=-\left(\frac{v^{2}+v+1}{v-1}\right) \end{gathered}$
$\begin{aligned} &d v\left(\frac{v-1}{v^{2}+v+1}\right)=-\frac{d x}{x} \\ &\int \frac{(v-1)}{v^{2}+v+1} d v=-\int \frac{d x}{x} \\ &\int \frac{(v-1)}{v^{2}+v+1} d v=-\log |x|+C \end{aligned}$
$\begin{aligned} &\text { Use } v^{2}+v+1=v^{2}+\frac{1}{2} \times 2 v+\frac{\left(1^{2}\right)}{\left(2^{2}\right)}+1-\frac{\left(1^{2}\right)}{2^{2}} \\ &\left(v+\frac{1}{2}\right)^{2}+1-\frac{1}{4} \\ &\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4} \end{aligned}$
$\begin{aligned} &\text { Put } v^{2}+v+1=\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4} \\ &v-1=v+\frac{1}{2}-\frac{1}{2}-1=\left(v+\frac{1}{2}\right)-\frac{3}{2} \end{aligned}$
$\begin{aligned} &\int \frac{\left(v+\frac{1}{2}\right)-\frac{3}{2}}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v=-\log |x|+C \\ &\int \frac{v+\frac{1}{2}}{\left(v+\frac{1}{4}\right)^{2}+\frac{3}{4}} d v-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v=-\log |x|+C \end{aligned}$
So, our equation become
$I_{1}-\frac{3}{2} I_{2}=-\log |x|+c$ ....(ii)
Solving $I_{1}$ ,
$\begin{aligned} &I_{1}=\int \frac{v+\frac{1}{2}}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} d v \\ &\text { Put }\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}=t \end{aligned}$
Diff.w.r.t.v
$\begin{aligned} &\frac{d\left(\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}\right)}{d v}=\frac{d t}{d v} \\ &2\left(v+\frac{1}{2}\right)=\frac{d t}{d v} \\ &d v=\frac{d t}{2\left(v+\frac{1}{2}\right)} \end{aligned}$
Put value of v and dv in $I_{1}$
$\begin{aligned} &I_{1}=\int \frac{v+\frac{1}{2}}{t} \times \frac{d t}{2\left(v+\frac{1}{2}\right)} \\ &=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t| \\ &=\frac{1}{2} \log \left|\left(v+\frac{1}{2}\right)^{2}+\frac{3}{2}\right| \\ &\frac{1}{2} \log \left|v^{2}+v+1\right| \end{aligned}$
Solving $I_{1}$
$\begin{aligned} &I_{2}=\int \frac{d v}{\left(v+\frac{1}{2}\right)^{2}+\frac{3}{4}} \\ &\quad=\int \frac{d v}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &\text { Using } \int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \end{aligned}$
$\begin{aligned} &\text { Where } \mathrm{x}=v+\frac{1}{2} \text { and } a=\frac{\sqrt{3}}{2} \\ &=\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1} \frac{\left(v+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}} \\ &=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right) \end{aligned}$
From (ii)
$\begin{aligned} &I_{1}-\frac{3}{2} I_{2}=-\log |x|+C \\ &\frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+C \\ &\frac{1}{2} \log \left|v^{2}+v+1\right|-\sqrt{3} \tan ^{-1} \frac{2 v+1}{\sqrt{3}}=-\log |x|+C \end{aligned}$
Replace v by $\frac{y}{x}$
$\begin{aligned} &\frac{1}{2} \log \left|\left(\frac{y}{x}\right)^{2}+\frac{y}{x}+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 \frac{y}{x}+1}{\sqrt{3}}\right)=-\log |x|+C \\ &\frac{1}{2} \log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\log |x|=\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C \end{aligned}$
Multiply both side by 2
$\begin{aligned} &\log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\log |x|^{2}=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 C \\ &\text { Put } 2 c=c \\ &\log \left(\frac{x^{2} y^{2}}{x^{2}}+\frac{x^{2} y}{x}+x^{2}\right)=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+C \\ &\log \left|y^{2}+x y+x^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+C \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (ii)

Answer : $\sin \frac{y}{x}=\log |x|+c$
Hint : You must know the rules of solving differential equation and integration
Given : $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$
Solution : given differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$
$\frac{d y}{d x}=\frac{y \cos \frac{y}{x}+x}{x \cos \frac{y}{x}} \quad \ldots .(i)$
It is a homogeneous differential equation
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
Eqn (i) becomes
$\begin{aligned} v+x \frac{d v}{d x} &=\frac{v x \cos v+x}{x \cos v} \\ x \frac{d v}{d x} &=\frac{v \cos v+1}{\cos v}-v \\ x \frac{d v}{d x} &=\frac{v \cos v+1-v \cos v}{\cos v} \end{aligned}$
$\begin{aligned} &x \frac{d v}{d x}=\frac{1}{\cos v} \\ &\cos v d v=\frac{d x}{x} \end{aligned}$
Integrating both sides
$\begin{aligned} &\sin v=\log |x|+c \\ &\sin \frac{y}{x}=\log |x|+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (iii)

Answer : $c y=\log \left|\frac{y}{x}\right|-1$
Hint : You must know the rules of solving differential equation and integration
Given : $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$
Solution : $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$
$\begin{aligned} &d y\left(x \log \left(\frac{y}{x}\right)-2 x\right)=-y d x\\ &\frac{d y}{d x}=\frac{-y}{x \log \left(\frac{y}{x}\right)-2 x}\\ &\frac{d y}{d x}=\frac{-y}{-x\left(2-\log \left(\frac{y}{x}\right)\right)}\\ &\frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \end{aligned}$
Step : 2
Putting $f(x, y)=\frac{d y}{d x} \text { and finding }(f(\lambda x, \lambda x))$
$\begin{gathered} f(x, y)=\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ \qquad \begin{array}{c} f(\lambda x, \lambda y)=\frac{\frac{\lambda y}{\lambda x}}{2-\log \left(\frac{\lambda y}{\lambda x}\right)} \\ =\frac{\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)} \\ f(\lambda x, \lambda y)=f(x, y) \end{array} \end{gathered}$
Step : 3
Solve by $y=vx$
Put $y=vx$
Diff w.r.t.x
$\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v \frac{d x}{d x} \\ &\frac{d y}{d x}=x \frac{d v}{d x}+v \end{aligned}$
Put value of $\frac{d y}{d x} \text { and } y=v x \text { in eqn(i) }$
$\begin{aligned} &\frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \frac{y}{x}} \\ &x \frac{d v}{d x}=\frac{v}{2-\log v}-v \\ &x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v} \end{aligned}$
$\begin{aligned} &x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v} \\ &\frac{2-\log v}{v \log v-v} d v=\frac{d x}{x} \end{aligned}$
Integrate both sides
$\begin{aligned} &\int \frac{2-\log v}{v \log v-v} d v=\int \frac{d x}{x} \\ &\int \frac{1+1-\log v}{v(\log v-1)} d v=\log x+\log c \\ &\int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\log x+\log c \\ &\int \frac{d v}{v(\log v-1)}-\log v=\log x+\log c \end{aligned}$
Put $t=\log v-1$
$dt = \frac{1}{v}dv$
So our equation
$\int \frac{d t}{t}-\log v=\log x+\log c$
Put value of t
$\begin{aligned} &\log (\log v-1)-\log v=\log x+\log c \\ &\log (\log v-1)=\log x+\log c+\log v \\ &\log (\log v-1)=\log (x c v) \\ &\text { put } v=\frac{y}{x} \end{aligned}$
$\begin{aligned} &\log \left(\log \frac{y}{x}-1\right)=\log x c \frac{y}{x} \\ &\log \left(\log \frac{y}{x}-1\right)=\log c y \\ &\log \frac{y}{x}-1=c y \\ &\Rightarrow c y=\log \frac{y}{x}-1 \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (iv)

Answer : $\frac{1}{2}(\sin x-\cos x)+c e^{x}$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}-y=\cos x$
Solution : differential equation is in form of
$\begin{gathered} \frac{d y}{d x}+P y=Q \\ P=-1 \quad, Q=\cos x \\ \text { I.F }=e^{\int P d x}=e^{\int-1 d x}=e^{-x} \end{gathered}$
Solution is
$\begin{aligned} &y \times I F=\int Q \times I F d x+c\\ &y e^{-x}=\int e^{-x} \cos x+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)\\ &\text { let } I=\int e^{-x} \cos x d x \end{aligned}$
Integrate by $\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left\{f^{\prime}(x) \int g(x) d x\right\} d x$
$\begin{gathered} \text { Take } f(x)=\cos x \text { and } g(x)=e^{-x} \\ I=\cos x \int e^{-x} d x-\int\left\{(-\sin x) \int e^{-x} d x\right\} d x \\ I=-e^{-x} \cos x-\int-\sin x\left(-e^{-x}\right) d x \end{gathered}$
$\begin{aligned} &I=-e^{-x} \cos x-\int e^{-x} \sin x d x \\ &I=-e^{-x} \cos x-\left\{\sin x \int e^{-x} d x-\int\left(\cos x \int e^{-x} d x\right) d x\right\} \end{aligned}$
$\begin{aligned} &I=-e^{-x} \cos x-\left(-e^{-x} \sin x\right)+\int-e^{-x} \cos x d x \\ &I=-e^{-x} \cos x+e^{-x} \sin x-\int e^{-x} \cos x d x \\ &I=e^{-x}(\sin x-\cos x)-I \\ &2 I=e^{-x}(\sin x-\cos x) \\ &I=\frac{e^{-x}}{2}(\sin x-\cos x) \end{aligned}$
From (i)
$\begin{aligned} &y e^{-x}=\int e^{-x} \cos x+c \\ &y e^{-x}=\frac{e^{-x}}{2}(\sin x-\cos x)+c \\ &y=\frac{1}{2}(\sin x-\cos x)+c e^{x} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (v)

Answer : $y=\frac{x^{2}}{4}+c x^{-2}$
Hint : You must know the rules of solving differential equation and integration
Given : $x \frac{d y}{d x}+2 y=x^{2} \quad, x \neq 0$
Solution : $x \frac{d y}{d x}+2 y=x^{2}$
Divide both sides by x
$\begin{gathered} \frac{x}{x} \frac{d y}{d x}+\frac{2 y}{x}=\frac{x^{2}}{x} \\ \frac{d y}{d x}+\frac{2 y}{x}=x \end{gathered}$
Differentiate equation in the form ,
$\frac{dy}{dx}+Py=Q$
Where, $P=\frac{2}{x}, Q=x$
$\text { I.F }=e^{\int \frac{2}{x} d x}$
$\begin{array}{lc} \text { I. } F=e^{2 \log x} &\; \; \; \; \; \; \; \; \; \; \left(\log x=\log x^{n}\right) \\ \text { I. } F=e^{\log x^{2}} & \left(e^{\log x}=x\right) \end{array}$
$\begin{array}{lc} \text { I. } F=x^{2} \end{array}$
Solution of differential equation is
$\begin{gathered} y \times I . F=\int Q \times I . F d x+C \\ y x^{2}=\int x \times x^{2} d x+C \\ y x^{2}=\int x^{3} d x+C \end{gathered}$
$\begin{aligned} x^{2} y &=\frac{x^{4}}{4}+C \\ y &=\frac{x^{2}}{4}+C x^{-2} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (vi)

Answer : $y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}+2 y=\sin x$
Solution : put in form $\frac{d y}{d x}+P y=Q$
$\frac{d y}{d x}+2 y=\sin x$ ...(i)
Step : find P and Q
Compare (i) with $\frac{d y}{d x}+P y=Q$
$P=2 \; \; \; \; \; \; \; \; \; \; \; Q=\sin x$
Find integration factor I.F
$\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\qquad \text { I.F }=e^{\int 2 d x}=e^{2 x} \\ &\text { So } I . F=e^{2 x} \end{aligned}$
Step : 4
$y \times I . F=\int Q \times I . F d x+C$
Putting value
$\begin{aligned} &y e^{2 x}=\int \sin x e^{2 x} d x+c\\ &\text { Let } I=\int \sin x e^{2 x} d x \end{aligned}$
Solving I
$\begin{aligned} &I=\int \sin x e^{2 x} d x \\ &=\sin x \int e^{2 x} d x-\int\left[\frac{d}{d x} \sin x \int e^{2 x} d x\right] d x \\ &=\sin x \frac{e^{2 x}}{2}-\int \cos x \frac{e^{2 x}}{2} d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \sin x e^{2 x}-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left\{-\sin x \int \frac{e^{2 x}}{2} d x\right\} d x\right] \\ &=\frac{1}{2} \sin x e^{2 x}-\frac{1}{2}\left(\cos x \frac{e^{2 x}}{2}\right)-\frac{1}{4} \int \sin x e^{2 x} d x \\ &I=\frac{1}{2} \sin x e^{2 x}-\frac{1}{4} \cos x e^{2 x}-\frac{1}{4} I \end{aligned}$
$\begin{aligned} &I+\frac{1}{4} I=\frac{1}{4}\left(2 \sin x e^{2 x}-\cos x e^{2 x}\right) \\ &\frac{5 I}{4}=\frac{e^{2 x}}{4}(2 \sin x-\cos x) \\ &I=\frac{e^{2 x}}{5}(2 \sin x-\cos x) \end{aligned}$
Substituting I in eq (ii)
$y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+c$

Differential Equation Exercise Revision Exercise (RE) Question 66 (vii)

Answer : $y=e^{-2 x}+c e^{-3 x}$
Hint : You must know the rules of solving differential equation and integration
Given : $\frac{d y}{d x}+3 y=e^{-2 x}$
Solution : put in form $\frac{d y}{d x}+P y=Q$
$\frac{d y}{d x}+3 y=e^{-2 x}$
Step : 2
Find P and Q by comparing , $P=3, Q=e^{-2 x}$
Step : 3
Find integrating factor
$\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\text { I.F }=e^{\int 3 d x} \\ &I . F=e^{3 x} \end{aligned}$
Step : 4
Solution of equation
$y \times I . F=\int Q \times I . F d x+c$
Putting values
$\begin{gathered} y \times e^{3 x}=\int e^{-2 x+3 x} d x+c \\ y e^{3 x}=\int e^{x} d x+c \\ y e^{3 x}=e^{x} d x+c \end{gathered}$
divide by $e^{3x}$
$y=e^{-2 x}+c e^{-3 x}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (viii)

Answer : $x y=\frac{x^{4}}{4}+c$
Hint : : integrate by applying integration of $y^{n}$
Given : $\frac{d y}{d x}+\frac{y}{x}=x^{2}$
Solution : $\frac{d y}{d x}+\frac{y}{x}=x^{2}$
Differential equation is in the form of
$\frac{d y}{d x}+Py=Q$
$P=\frac{1}{x}$ , $Q=x^{2}$
Putting I.F
$\begin{aligned} &\text { I.F }=e^{\int P d x} \\ &\text { I.F }=e^{\int \frac{1}{x} d x} \\ &\text { I.F }=e^{\log x} \\ &\text { I.F }=x \end{aligned}$
Solution is
$\begin{aligned} &y \times I . F=\int(Q \times I . F) d x+c \\ &y x=\int x^{2} \times x d x+c \\ &y x=\int x^{3} d x+c \\ &y x=\frac{x^{4}}{4}+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (ix)

Answer : $y(\sec x+\tan x)=\sec x+\tan x-x+c$
Hint : integrate by applying integration of sec x and tan x
Given : $\frac{d y}{d x}+\sec x(y)=\tan x$
Solution : differential equation is of the form
$\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &\text { Where }_{L} P=\sec x, \quad Q=\tan x \end{aligned}$
Finding integrating factor
$\begin{aligned} &I . F=e^{\int P d x} \\ &I . F=e^{\int \sec x d x} \\ &I \cdot F=e^{\log |\sec x+\tan x|} \\ &\text { I. } F=\sec x+\tan x \end{aligned}$
Solution is
$\begin{gathered} y \times I . F=\int(Q \times I . F) d x+c \\ y(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx+c \\ y(\sec x+\tan x)=\int \tan x \sec x d x+\int \tan ^{2} d x+c \end{gathered}$
$\begin{aligned} &y(\sec x+\tan x)=\int \tan x \sec x d x+\int\left(\sec ^{2} x-1\right) d x+c \\ &y(\sec x+\tan x)=\sec x+\tan x-x+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (x)

Answer : $y=\frac{x^{2}}{16}(4 \log |x|-1)+c x^{-2}$
Hint : integrate by applying integration of $x^{n}$ and $log\; x$
Given : $x \frac{d y}{d x}+2 y=x^{2} \log x$
Solution : convert the given differential equation is of the form $\frac{d y}{d x}+P y=Q$
$x \frac{d y}{d x}+2 y=x^{2} \log x$
Divide both sides by x
$\begin{aligned} &\frac{d y}{d x}+2 \frac{y}{x}=x \log x \\ &\text { Now } P=\frac{2}{x} \quad, Q=x \log x \end{aligned}$
Differentiate
$\begin{aligned} &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=e^{\int \frac{2}{x} d x} \\ &\text { I.F }=e^{2 \int \frac{1}{x} d x} \\ &\text { I. } F=e^{2 \log x} \\ &\text { I. } F=e^{\log x^{2}}=x^{2} \end{aligned}$
Solution is
$\begin{gathered} y \times \text { I.F }=\int(Q \times I . F) d x+c \\ y x^{2}=\int x \log x \times x^{2} d x+c \\ y x^{2}=\int \log x \times x^{3}+c \end{gathered}$
$\begin{aligned} &\text { Use } \int f(x) \times g(x) d x=f(x) \int g(x) d x-\int\left\{f^{\prime}(x) \int g(x) d x\right\} d x \\ &y x^{2}=\log x \int x^{3} d x-\int\left\{\frac{d}{d x} \log x \int x^{3} d x\right\} d x \\ &y x^{2}=\log x\left(\frac{x^{4}}{4}\right)-\int \frac{1}{x}\left(\frac{x^{4}}{4}\right) d x+c \end{aligned}$
$\begin{aligned} &y x^{2}=\frac{\log x \times x^{4}}{4}-\int \frac{x^{3}}{4} d x+c \\ &y x^{2}=\frac{x^{4} \log x}{4}-\frac{x^{4}}{16}+c \end{aligned}$
$\begin{aligned} &y=\frac{x^{4} \log x}{4 x^{2}}-\frac{x^{4}}{16 x^{2}}+\frac{c}{x^{2}} \\ &y=\frac{x^{2} \log |x|}{4}-\frac{x^{2}}{16}+c x^{-2} \\ &y=\frac{x^{2}}{16}(4 \log |x|-1)+c x^{-2} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (xi)

Answer : $y \log x=-\frac{2}{x}(1+\log x)+c$
Hint : you integrate by integrating $x^{n}$
Given : $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
Solution : $\text { put in form } \frac{d y}{d x}+P y=Q$
$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
Divide bot side by $(x \log x)$
$\begin{aligned} &\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \log x \times \frac{1}{x \log x}\\ &\frac{d y}{d x}+\left(\frac{1}{x \log x}\right) y=\frac{2}{x^{2}} \end{aligned}$ ...(i)
By comparing (i) with $\frac{d y}{d x}+P y=Q$
$\begin{aligned} &P=\frac{1}{x \log x} \quad \text { and } Q=\frac{2}{x^{2}} \\ &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=\int e^{\frac{1}{x \log x} d x} \\ &\text { Let } t=\log x \end{aligned}$
$\begin{aligned} &d t=\frac{1}{x} d x \\ &I . F=e^{\int \frac{1}{t} d t} \\ &I \cdot F=e^{\log |t|} \\ &\text { I.F }=|t|=\log |x| \end{aligned}$
$\begin{gathered} y \times \text { I.F }=\int Q \times \text { I.F } d x+c \\ y \times \log x=\int \frac{2}{x^{2}} \log x d x+c \end{gathered}$ ....(ii)
$\text { Let } I=2 \int \log x \cdot x^{-2} d x$
$\begin{aligned} &I=2\left[\log x \int x^{-2} d x-\int\left\{\frac{1}{x} \int x^{-2} d x\right\} d x\right] \\ &I=2\left[\log x \frac{x^{-1}}{-1}-\int \frac{1}{x} \frac{x^{-1}}{-1} d x\right] \end{aligned}$
$\begin{aligned} &=2\left(-\log x \frac{1}{x}+\int \frac{1}{x^{2}} d x\right) \\ &I=2\left(\frac{-1}{x} \log x-\frac{1}{x}\right)=\frac{-2}{x}(1+\log x) \end{aligned}$
Now enq (ii) becomes
$\begin{aligned} &y \log x=I+C \\ &y \log x=\frac{-2}{x}(1+\log x)+c \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (xii)

Answer : $y=\left(1+x^{2}\right)^{-1} \log |\sin x|+c\left(1+x^{2}\right)^{-1}$
Hint : integrate by applying integration of $x^{n}$
Given : $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x$
Solution : given equation
$\left(1+x^{2}\right) d y+2 x y d x=\cot x d x$
Divide both sides by dx
$\begin{gathered} \left(1+x^{2}\right) \frac{d y}{d x}+2 x y \frac{d x}{d x}=\cot x \frac{d x}{d x} \\ \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\cot x \end{gathered}$
Divide by $(1+x^{2})$
$\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$
Comparing above equation with $\frac{d y}{d x}+P y=Q$
$\begin{aligned} &\quad P=\frac{2 x}{1+x^{2}} \quad, Q=\frac{\cot x}{1+x^{2}} \\ &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=e^{\int \frac{2 x}{1+x^{2}} d x} \\ &\text { Let } t=1+x^{2} \end{aligned}$
$\begin{gathered} d t=2 x d x \\ I . F=e^{\int \frac{d t}{t}}=e^{\log t}=t=\left(1+x^{2}\right) \end{gathered}$
$\begin{aligned} y \times I . F &=\int Q \times I . F d x+c \\ y\left(1+x^{2}\right) &=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+c \\ y\left(1+x^{2}\right) &=\int \cot x d x+c \end{aligned}$
$\begin{aligned} &y\left(1+x^{2}\right)=\log |\sin x|+c \\ &\text { Divide } 1+x^{2} \\ &y=\left(1+x^{2}\right)^{-1} \log |\sin x|+c\left(1+x^{2}\right)^{-1} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (xiii)

Answer : $x+y+1=ce^{y}$
Hint : you integrate by applying integration of $x^{n}$
Given : $(x+y) \frac{d y}{d x}=1$
Solution : Put in the form of $\frac{dy}{dx}+Py=Q$
$(x+y) \frac{d y}{d x}=1$
Divide by $x+y$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{x+y} \\ \end{aligned}$
$\frac{d x}{d y}=x+y$
$\frac{d x}{d y}+(-x)=y$ .....(i)
Find $P_{1}$ and $Q_{1}$
Comparing (i)
$\begin{aligned} &\frac{d y}{d x}+P_{1}=Q_{1} \\ &P_{1}=-1 \quad, Q_{1}=y \end{aligned}$
Find I.F
$\begin{aligned} &\text { I.F }=e^{\int P_{1} d y} \\ \end{aligned}$
$\begin{aligned} &\quad=e^{\int-1 d y}=e^{-y} \\ \end{aligned}$
$x \times I . F=\int Q_{1} \times I . F d y+c$
Putting value $x e^{-y}=\int y \times e^{-y} d y+c$ ......(ii)
$\begin{aligned} &\text { Let } I=\int y e^{-y} d y \\ \end{aligned}$
$\begin{aligned} &=y \int e^{-y} d y-\int\left(\frac{d}{d y} y \int e^{-y} d y\right) d y \\ \end{aligned}$
$\begin{aligned} &=y \frac{e^{-y}}{-1}-\int \frac{e^{-y}}{-1} d y \\ \end{aligned}$
$\begin{aligned} &=-y e^{-y}+\int e^{-y} d y \\ \end{aligned}$
$\begin{aligned} &I=-y e^{-y}-e^{-y} \end{aligned}$
Put value of I in (ii)
$\begin{aligned} &x e^{-y}=\int y e^{-y} d y+c \\ &x e^{-y}=-y e^{-y}-e^{-y}+c \end{aligned}$
Divide by $e^{-y}$
$\begin{aligned} &x=-y-1+c e^{y} \\ &x+y+1=c e^{y} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (xiv)

Answer : $x=\frac{y^{3}}{3}+\frac{c}{y}$
Hint : integrate by applying integration of $x^{n}$
Given : $y \; d x+\left(x-y^{2}\right) d y=0$
Solution : $y \; d x+\left(x-y^{2}\right) d y=0$
$\frac{d y}{d x}=\frac{-y}{x-y^{2}}$
This is not in the form of $\frac{d y}{d x}+Py=Q$
$\begin{aligned} &\frac{d x}{d y}=\frac{y^{2}-x}{y} \\ \end{aligned}$
$\begin{aligned} &\frac{d x}{d y}=y-\frac{x}{y} \\ \end{aligned}$
$\begin{aligned} &\frac{d x}{d y}+\frac{x}{y}=y \end{aligned}$ .....(i)
Find P and Q Where $P=\frac{1}{y} \quad, Q=y$
Find I.F
$\begin{aligned} \text { I.F } &=e^{\int P d y} \\ &=e^{\int \frac{1}{y} d y}=e^{\log y}=y \end{aligned}$
Solution will be
$\begin{aligned} &x(I . F)=\int(Q \times I . F) d y+c \\ &x y=\int y \times y d y+c \\ &x y=\int y^{2} d y+c \end{aligned}$
$\begin{aligned} x y &=\frac{y^{3}}{3}+c \\ x &=\frac{y^{2}}{3}+\frac{c}{y} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 66 (xv)

Answer : $x=3 y^{2}+c y$

Hint : integrate by applying integration of $x^{n}$

Given : $\left(x+3 y^{2}\right) \frac{d y}{d x}=y$

Solution : $\left(x+3 y^{2}\right) \frac{d y}{d x}=y$

This is not in the foem of $\frac{d x}{d y}+Px=Q$

Where $P_{1}=\frac{-1}{y} \& Q=3 y$
Step : 3 find integration factor
$\begin{aligned} &\text { I. } F=e^{\int P d y} \\ &\text { I. } F=e^{\int \frac{-1}{y} d y} \Rightarrow e^{-\log y}=e^{\log y^{-1}}=y^{-1}=\frac{1}{y} \end{aligned}$
Step : 4
Solution is
$\begin{aligned} x(I . F) &=\int Q \times I . F d y+c \\ x \frac{1}{y} &=\int 3 y \frac{1}{y} d y+c \\ \frac{x}{y} &=3 \int d y+c \end{aligned}$
$\begin{array}{r} \frac{x}{y}=3 y+c \\ \end{array}$
$\begin{array}{r} x=3 y^{2}+c y \end{array}$

Differential Equation Exercise Revision Exercise (RE) Question 67 (i)

Answer : $y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}$
Hint : Use the formula $\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c$
Given : $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} ; y=0 \text { when } x=1$
Solution : $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}}$
Divide by $1+x^{2}$
$\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^{2}}\right) y=\frac{1}{\left(1+x^{2}\right)^{2}}$
This is a linear differential equation of the form
Here, $P=\frac{2 x}{1+x^{2}} \quad ; Q=\frac{1}{\left(1+x^{2}\right)^{2}}$
The integrating factor of this differential equation is
$\text { I.F }=e^{\int P d x}=e^{\int \frac{2 x}{1+x^{2}} d x}$
We have, $\begin{aligned} &\int \frac{2 x}{1+x^{2}}=\log \left|1+x^{2}\right|+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{d x}{x}=\log |x|+c\right] \\ \end{aligned}$
Therefore, $\begin{aligned} &\text { I. } F=e^{\log \left|1+x^{2}\right|}=1+x^{2} \\ \end{aligned}$
$\begin{aligned} &\text { I. } F=1+x^{2} \end{aligned}$
Hence the solution of the differential equation is
$\begin{gathered} y(I . F)=\int(Q \times I . F) d x+c \\ y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}}\left(1+x^{2}\right) d x+c \\ \Rightarrow y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+c \end{gathered}$ ...(i)
We know , $\int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c$
Therefore, $y\left(1+x^{2}\right)=\tan ^{-1} x+c$ ....(ii)
Now $y=0$ when $x=1$
Therefore,
$\begin{aligned} 0\left(1+1^{2}\right) &=\tan ^{-1}(1)+c \\ \Rightarrow 0 &=\frac{\pi}{4}+c \Rightarrow c=-\frac{\pi}{4} \end{aligned}$
By (ii)
$y\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}$

Differential Equation Exercise Revision Exercise (RE) Question 67 (ii)

Answer : $\tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\frac{y^{2}}{x^{2}}}+1 \times x\right|=\frac{\pi}{4}+\frac{1}{2} \log 2$
Hint : using variable separable method and substituting the values
Given : $(x+y) d y+(x-y) d x=0 \quad, y=1 \text { when } x=1$
Solution :
$\begin{aligned} &(x+y) d y+(x-y) d x=0 \\ &\Rightarrow(x+y) d y=-(x-y) d x \\ &\Rightarrow(x+y) d y=(y-x) d x \\ &\Rightarrow \frac{d y}{d x}=\frac{y-x}{x+y}=\frac{-(x-y)}{x+y} \end{aligned}$
Now let ,
$\begin{aligned} &f(x, y)=\frac{d y}{d x}=\frac{-(x-y)}{x+y} \\ &\text { Finding } f(\lambda x, \lambda y) \end{aligned}$
$\begin{aligned} f(\lambda x, \lambda y) &=\frac{-(\lambda x-\lambda y)}{\lambda x+\lambda y} \\ &=\frac{-\lambda(x-y)}{\lambda(x+y)} \\ &=\frac{-(x-y)}{x+y} \\ &=\lambda^{0} f(x, y) \end{aligned}$
Therefore, $f(x,y)$ is a homogeneous function of degree 0 .
Putting $y=vx$
Diff w.r.t.x
$\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v\\ &\text { Putting value of } \frac{d y}{d x} \text { and } y=v x \text { in (i) } \end{aligned}$
$\begin{gathered} \frac{d y}{d x}=\frac{-(x-y)}{x+y} \\ v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x} \\ v+x \frac{d v}{d x}=\frac{-x(1-v)}{x(1+v)} \end{gathered}$
$\begin{gathered} =\frac{v-1}{1+v} \\ \Rightarrow x \frac{d v}{d x}=\frac{v-1}{1+v}-v \\ \Rightarrow x \frac{d v}{d x}=\frac{v-1-v(1+v)}{1+v} \end{gathered}$
$\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{-\left(1+v^{2}\right)}{1+v} \\ &\Rightarrow \frac{1+v}{1+v^{2}} d v=-\frac{d x}{x} \end{aligned}$
Integrating both sides
$\begin{aligned} &\int \frac{1+v}{1+v^{2}} d v=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{1}{1+v^{2}} d v+\int \frac{v}{v^{2}+1} d v=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\int \frac{v}{v^{2}+1} d v=-\log |x|+c \end{aligned}$
Putting $v^{2}+1=t$ in integral
$\begin{aligned} &2 v d v=d t \\ &v d v=\frac{d t}{2} \end{aligned}$
$\begin{aligned} &\Rightarrow \tan ^{-1} v+\int \frac{1}{t} \times \frac{d t}{2}=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\frac{1}{2} \log |t|=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\frac{1}{2} \log \left|v^{2}+1\right|=-\log |x|+c \end{aligned}$
Now again putting back the value of $v=\frac{y}{x}$
$\begin{aligned} &\tan ^{-1} \frac{y}{x}+\frac{1}{2} \log \left|\left(\frac{y}{x}\right)^{2}+1\right|+\log |x|=c \\ &\Rightarrow \tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\left(\frac{y}{x}\right)^{2}+1}\right|+\log |x|=c \\ &\Rightarrow \tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\left(\frac{y}{x}\right)^{2}+1} \times x\right|=c \end{aligned}$ ....(ii)
Now, $y = 1$ when $x = 1$
$\begin{aligned} &\text { Therefore } \tan ^{-1}\left(\frac{1}{1}\right)+\log \left|\sqrt{\left(\frac{1}{1}\right)^{2}+1} \times 1\right|=c \\ &\tan ^{-1}(1)+\log \sqrt{2}=c \\ &\frac{\pi}{4}+\frac{1}{2} \log 2=c \end{aligned}$
Put in (ii)
$\tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\frac{y^{2}}{x^{2}}+1} \times x\right|=\frac{\pi}{4}+\frac{1}{2} \log 2$

Differential Equation Exercise Revision Exercise (RE) Question 67 (iii)

Answer : $y+2 x=3 x^{2} y$
Hint : using variable separable method and substituting the values
Given : $x^{2} d y+\left(x y+y^{2}\right) d x=0, \quad y=1, \text { when } x=1$
Solution : $x^{2} d y+\left(x y+y^{2}\right) d x=0$
The differential equation can be written as
$\begin{aligned} &x^{2} d y=-\left(x y+y^{2}\right) d x\\ &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}$ ......(i)
Let $f(x, y)=\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}}$
Finding $f(\lambda \; x,\lambda \; y)$
$\begin{aligned} f(\lambda x, \lambda y) &=\frac{-\left(\lambda x . \lambda y+\lambda^{2} y^{2}\right)}{\lambda^{2} x^{2}}=\frac{-\lambda^{2}\left(x y+y^{2}\right)}{\lambda^{2} x^{2}} \\ &=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &=\lambda^{0}(f(x, y)) \\ f(x, y)=& \frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}$
Therefore , $f(x,y)$ is a homogenous function of degree zero.
Putting $y=vx$
Diff w.r.t.x
$\frac{d y}{d x}=x \frac{d v}{d x}+v$
Putting value of $\frac{d y}{d x} \text { and } y=v x \text { in }(i)$
$\begin{aligned} &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x(v x)+(v x)^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x^{2} v+x^{2} v^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-x^{2}\left(v+v^{2}\right)}{x^{2}} \end{aligned}$
$\begin{aligned} &x \frac{d v}{d x}=-v-v^{2}-v=-\left(v^{2}+2 v\right) \\ &\frac{d v}{v^{2}+2 v}=-\frac{d x}{x} \end{aligned}$
Integrating both sides we get ,
$\begin{aligned} &\int \frac{d v}{v^{2}+2 v}=-\int \frac{d x}{x} \\ \end{aligned}$
$\begin{aligned} &\int \frac{d v}{\left(v^{2}+2 v+1\right)-1}=-\log x+\log c \\ \end{aligned}$
$\begin{aligned} &\int \frac{d v}{(v+1)^{2}-1^{2}}=-\log x+\log c \\ \end{aligned}$
Using $\begin{aligned} & \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c \\ \end{aligned}$
$\begin{aligned} &\frac{1}{2} \log \frac{v+1-1}{v+1+1}=-\log x+\log c \end{aligned}$
$\begin{aligned} &\frac{1}{2} \log \frac{v}{v+2}=-\log x+\log c \end{aligned}$
$\begin{aligned} &\log \sqrt{\frac{v}{v+2}}+\log x=\log c \\ &\frac{x \sqrt{v}}{\sqrt{v+2}}=c \end{aligned}$
Now putting value of v i.e , y/x
$\begin{aligned} &\Rightarrow \frac{x \sqrt{\frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}$
$\begin{aligned} &\frac{\sqrt{x^{2} \times \frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}$
$\begin{aligned} &\frac{\sqrt{x y}}{\sqrt{\frac{y+2 x}{x}}}=c \\ \end{aligned}$
$\begin{aligned} &\frac{x \sqrt{y}}{\sqrt{y+2 x}}=c \\ \end{aligned}$
$\begin{aligned} &x \sqrt{y}=c \sqrt{y+2 x} \end{aligned}$
Squaring both sides
$\begin{aligned} &x^{2} y=c^{2}(y+2 x) \quad \; \; \; \; \; \; \; \; \; \ldots (ii)\\ &\text { Now } x=1, y=1\\ &\text { Therefore }\left(1^{2}\right)(1)=c^{2}(1+2(1)) \end{aligned}$
$\begin{aligned} &\Rightarrow 1=c^{2}(3) \\ &\Rightarrow c^{2}=\frac{1}{3} \end{aligned}$
Putting back in (ii)
$\begin{aligned} &\Rightarrow x^{2} y=\frac{1}{3}(y+2 x) \\ &3 x^{2} y=y+2 x \\ &y+2 x=3 x^{2} y \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 68

Answer : $y=x^{2}+\log\; \left | x \right |$
Hint : Using variable separable method
Given : $x \; d y=\left(2 x^{2}+1\right) d x \quad, x \neq 0$
Solution : $x \; d y=\left(2 x^{2}+1\right) d x \quad$
$\begin{aligned} &d y=\frac{\left(2 x^{2}+1\right)}{x} d x \\ &d y=\left(\frac{2 x^{2}}{x}+\frac{1}{x}\right) d x \\ &d y=\left(2 x+\frac{1}{x}\right) d x \end{aligned}$
Integrating both sides
$\begin{gathered} \int d y=\int\left(2 x+\frac{1}{x}\right) d x \\ \end{gathered}$ ....(i)
$\begin{gathered} \int d y=\int 2 x d x+\int \frac{1}{x} d x \\ \end{gathered}$
$\begin{gathered} y=\frac{2 x^{2}}{2}+\log |x|+c \\ \end{gathered}$
$\begin{gathered} y=x^{2}+\log |x|+c \end{gathered}$ .....(ii)
Since the curve passes through (1,1)
Putting $x=1, y=1$ in (ii)
$\begin{aligned} &1=(1)^{2}+\log (1)+c \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log 1=0] \\ &1=1+0+c=>c=0 \end{aligned}$
Put $c=0$ in (ii)
i.e , $\begin{gathered} y=x^{2}+\log |x|+c \\ \end{gathered}$
$y=x^{2}+\log |x|+0 \\$
$y=x^{2}+\log |x|$
Hence the equation of curve is
$y=x^{2}+\log\; \left | x \right |$

Differential Equation Exercise Revision Exercise (RE) Question 69

Answer : $y=\left(3 x^{2}+15\right)^{\frac{1}{3}}$
Hint : use variable separable method
Given : slope of the tangent to the curve at any $(x, y) \text { is } \frac{2 x}{y^{2}}$
Solution : Slope of tangent $=\frac{dy}{dx}$
$\begin{aligned} \frac{d y}{d x} &=\frac{2 x}{y^{2}} \\ \Rightarrow \quad y^{2} d y &=2 x d x \end{aligned}$
Integrating both sides
$\begin{gathered} \int y^{2} d y=\int 2 x d x \\ \end{gathered}$
$\begin{gathered} \frac{y^{3}}{3}=2 \frac{x^{2}}{2}+c \\ \end{gathered}$
$\begin{gathered} \frac{y^{3}}{3}=x^{2}+c \\ \end{gathered}$
$\begin{gathered} y^{3}=3 x^{2}+3 c \end{gathered}$
$\begin{aligned} &y^{3}=3 x^{2}+c_{1} &.....\text { (i) where } c_{1}=3 c \end{aligned}$
Given that equation passes through (-2, 3)
Putting $x=-2, y=3$ in (i)
$\begin{aligned} &y^{3}=3 x^{2}+c_{1} \\ &3^{3}=3(-2)^{2}+c_{1} \\ &27=3 \times 4+c_{1} \quad \Rightarrow c_{1}=27-12=15 \end{aligned}$
Putting $C_{1}$ in (i)
$\begin{aligned} &y^{3}=3 x^{2}+15 \\ &y=\left(3 x^{2}+15\right)^{\frac{1}{3}} \end{aligned}$ is the perticular solution of the equation

Differential Equation Exercise Revision Exercise (RE) Question 70

Answer : $2 y-1=e^{x}(\sin x-\cos x)$
Hint : using integration by parts
Given : $\frac{d y}{d x}=e^{x} \sin x$
Solution : $\frac{d y}{d x}=e^{x} \sin x$
$d y=e^{x} \sin x d x$
Integrating both sides
$\begin{aligned} &\int d y=\int e^{x} \sin x d x\\ &y=\int e^{x} \sin x d x \end{aligned}$ ....(i)
Using integration by parts
Let $\begin{aligned} &I=\int e^{x} \sin x d x=\sin x e^{x}-\int \cos x e^{x} d x \\ \end{aligned}$
$\begin{aligned} &\qquad=\sin x e^{x}-\left[\cos x e^{x}-\int(-\sin x) e^{x} d x\right] \\ \end{aligned}$
$\begin{aligned} &I=\sin x e^{x}-\cos x e^{x}-\int \sin x e^{x} d x \\ \end{aligned}$
$\begin{aligned} &I=\sin x e^{x}-\cos x e^{x}-I \\ \end{aligned}$
$\begin{aligned} &2 I=\sin x e^{x}-\cos x e^{x} \\ \end{aligned}$
$\begin{aligned} &I=\frac{e^{x}(\sin x-\cos x)}{2} \end{aligned}$
Put in (i), we get,
$\begin{gathered} y=\frac{e^{x}(\sin x-\cos x)}{2} \\ 2 y=e^{x}(\sin x-\cos x) \\ \qquad \begin{array}{r} y=\frac{1}{2} e^{x}(\sin x-\cos x)+c \end{array} \end{gathered}$ ....(ii)

Given curve passes through (0,0)

Putting x = 0 , y = 0 in equation

Therefore , $\begin{aligned} &0=\frac{1}{2} e^{0}(\sin 0-\cos 0)+c \\ \end{aligned}$

$\begin{aligned} &0=\frac{1}{2}(1)(0-1)+c \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{0}=1\right] \\ \end{aligned}$

$\begin{aligned} &0=\frac{-1}{2}+c \end{aligned}$

$c=\frac{1}{2}$

Putting value of C in (ii)

$\begin{gathered} y=\frac{1}{2} e^{x}(\sin x-\cos x)+\frac{1}{2} \\ y-\frac{1}{2}=\frac{1}{2} e^{x}(\sin x-\cos x) \\ \frac{2 y-1}{2}=\frac{e^{x}}{2}(\sin x-\cos x) \\ 2 y-1=e^{x}(\sin x-\cos x) \end{gathered}$

Differential Equation Exercise Revision Exercise (RE) Question 71

Answer: $y+3=(x+4)^{2}$
Given:
At any point $(x , y)$ of a curve , the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$
Hint:
Using variable separable method and substituting the values.
Explanation:
Slope of the tangent to the curve $=\frac{dy}{dx}$
Slope of line segment joining $(x , y)$ and $(-4,-3)$
$\begin{aligned} &=\frac{y_{2}-y_{1}}{-4+x} \\ \end{aligned}$
$\begin{aligned} &=\frac{-3-y}{-4-x} \end{aligned}$
$\begin{aligned} &=\frac{-(y+3)}{-(x+4} \\ \end{aligned}$
$\begin{aligned} &=\frac{y+3}{x+4} \end{aligned}$
Given at point $(x , y)$ . Slope of tangent is twice of line segment
$\begin{aligned} &\frac{d y}{d x}=2\left(\frac{y+3}{x+4}\right)^{2} \\ &\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4} \end{aligned}$
Integrating both sides.
$\begin{aligned} &\int \frac{d y}{y+3}=2 \int \frac{d x}{x+4} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow \log |y+3|=2 \log |x+4|+\log c \\ \end{aligned}$
$\begin{aligned} &\Rightarrow \log |y+3|=\log |x+4|^{2}+\log c\left[\therefore a \log x=\log x^{2}\right] \\ \end{aligned}$
$\begin{aligned} &\Rightarrow \log (y+3)-\log (x+4)^{2}=\log c \\ \end{aligned}$
$\begin{aligned} &\log \frac{(y+3)}{(x+4)^{2}}=\log c \\ \end{aligned}$ $\begin{aligned} &\Rightarrow \frac{y+3}{(x+4)^{2}}=c-(1) \end{aligned}$

Since curve passes through (-2,1)
Put $x=-2 \; and\; y=1$ in (1)

$\begin{gathered} \frac{1+3}{(-2+4)^{2}}=C \\ \end{gathered}$

$\begin{gathered} \Rightarrow C=\frac{4}{(2)^{2}} \\ \end{gathered}$

$\begin{gathered} =\frac{4}{4} \\ =1 \\ \Rightarrow C=1 \end{gathered}$

Put value of c in eq (1)

$\begin{aligned} &\frac{y+3}{(x+4)^{2}}=1 \\ &\Rightarrow y+3=(x+4)^{2} \end{aligned}$
Hence the equation of the curve is
$y+3=(x+4)^{2}$

Differential Equation Exercise Revision Exercise (RE) Question 72

Answer : $x^{2}-y^{2}=c x$
Given : $\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}$
Hint:
Using variable separable method.
Explanation:
We know that the slope of the tangent at $(x,y)$ of a Curve is $\frac{dy}{dx}$

Given slope of tangent at $(x,y)$ is $\frac{x^{2}+y^{2}}{2 x y}$
$\therefore \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}-(1)$
$\begin{aligned} &\text { Put } \frac{d y}{d x}=f(x, y) \Rightarrow F(x, y)=\frac{x^{2}+y^{2}}{2 x y} \\ &\text { Finding } f(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{2(\lambda x)(\lambda y)} \\ \end{aligned}$
$\begin{aligned} &=\frac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{2 \lambda^{2} x y} \end{aligned}$
$\begin{aligned} &=\frac{\lambda^{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)}{\lambda^{2} \cdot 2 \mathrm{xy}} \\ \end{aligned}$
$\begin{aligned} &=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{xy}} \\ \end{aligned}$
$\begin{aligned} &=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}$
$\begin{aligned} &\text { So, } \mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}$
$\begin{aligned} &=\lambda^{\circ} \mathrm{F}(\mathrm{x}, \mathrm{y}) \end{aligned}$
So $\mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})$ is homogeneous function of degree 0
$\therefore$ Given equation is homogeneous differential equation
Now Put $y =vx$ in (1)
Differentiate with respect to x
$\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v\\ &\text { Putting } \frac{\mathrm{dy}}{\mathrm{dx}} \text { value in }(1) \text { and } \mathrm{y}=\mathrm{vx} \end{aligned}$
$\begin{aligned} &x \frac{d v}{d x}+v=\frac{x^{2}+(v x)^{2}}{2 x(v x)}=\frac{x^{2}+v^{2} x^{2}}{2 v x^{2}}=\frac{1+v^{2}}{2 v} \\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v \Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}-2 v^{2}}{2 v} \end{aligned}$
$\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \\ &\Rightarrow \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \cdot \frac{1}{x} \Rightarrow \frac{2 v}{1-v^{2}} d v=\frac{d x}{x} \end{aligned}$
$\begin{aligned} &\Rightarrow-\frac{2}{v^{2}-1} d v=\frac{d x}{x} \\ &\Rightarrow \frac{2 v d v}{v^{2}-1}=-\frac{d x}{x} \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{2 \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{~d} \mathrm{v}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\log |\mathrm{x}|+\mathrm{c} \end{aligned}$
Solving $\int \frac{2v}{v^{2}-1}dv$
Put $v^{2}-1=t$
Differentiate with respect to v
$\begin{gathered} 2 v d v=d t \\ d v=\frac{d t}{2 v} \\ =\int \frac{2 v}{t} \cdot \frac{d t}{2 v} \\ =\int \frac{d t}{t}=\log 1+1 \end{gathered}$
Putting back
$\begin{aligned} &\mathrm{t}=\mathrm{v}^{2}-1 \\ &\Rightarrow \log \left|\mathrm{v}^{2}-1\right| \end{aligned}$
$\therefore$ By (2)
$\begin{aligned} &\log \left|\mathrm{v}^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{c}_{1}\\ &\text { Putting }\\ &\mathrm{vx}=\mathrm{y} \text { or } \end{aligned}$
$\begin{aligned} &\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}} \\ &\log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|+\log |\mathrm{x}| \\ &=\mathrm{C}_{1} \end{aligned}$
$\begin{aligned} &\Rightarrow \log \left|\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right] \dot{\mathrm{x}}\right| \\ &=\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left[\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right] \mathrm{x}\right|=\mathrm{C}_{1} \end{aligned}$
Putting $C_{1}=\log\; C_{1}$
$\log \left|\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right) \mathrm{x}\right|=\log \mathrm{C}_{1}$
Remove log of both sides.
$\begin{aligned} &\\\left(\frac{y^{2}}{x^{2}}-1\right) x=C_{1} \\ \end{aligned}$
$\begin{aligned} \\\Rightarrow & \frac{x y^{2}}{x^{2}}-x=C_{1} \\ \end{aligned}$
$\begin{aligned} \\\Rightarrow \frac{y^{2}}{x}-x=C_{1} \end{aligned}$
$\begin{aligned} &\Rightarrow \mathrm{y}^{2}-\mathrm{x}^{2}=\mathrm{x} \mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=-\mathrm{C}_{1} \mathrm{x} \\ \\&\text { Put } \mathrm{C}=-\mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{cx} \end{aligned}$
Hence Proved

Differential Equation Exercise Revision Exercise (RE) Question 73

Answer: $-1+2e^{x^{2}/2}$
Given: The slope of the tangent to the curve at any point $(x , y)$ is equal to the sum of x coordinate and the product of x and y coordinate of that point
Hint: You must know about integrating factor
Explanation: Slope of the tangent to the curve at $(x , y)=\frac{dy}{dx}$
Given that,
Slope of the tangent to the curve at point $(x , y)$ is equal to the sum of x coordinate and the product of x and y coordinate of that point
So our equation becomes,
$\begin{aligned} &\frac{d y}{d x}=x+x y \\ \\&\Rightarrow \frac{d y}{d x}-x y=x \end{aligned}$
Differential equation is of the form
$\begin{aligned} &\frac{d y}{d x}+Py=Q \\ \end{aligned}$
Where
$\begin{aligned} &\mathrm{P}=-\mathrm{x} \text { or }\\ \\&\theta=x\\ \end{aligned}$
If
$\begin{aligned} &=e^{P d \int x}\\ \\&=e^{-\int x d x}\\ \\&=e^{-x^{2} / 2} \end{aligned}$
Solution is
$\begin{aligned} &\mathrm{y} \cdot(\mathrm{I} \cdot \mathrm{f} \cdot)=\int(\theta \times \mathrm{I} \cdot \mathrm{f} \cdot) \mathrm{d} \mathrm{x}+\mathrm{c} \\ &\mathrm{ye}^{-\frac{\mathrm{x}^{2}}{2}}=\int \mathrm{xe}^{-\frac{\mathrm{x}^{2}}{2}} \mathrm{dx}+\mathrm{c} \end{aligned}$
Putting ,
$\begin{aligned} &-\frac{x^{2}}{2}=t \\ \\&\Rightarrow \frac{-2 x}{2} d x=d t \\ \\&\Rightarrow x d x=a b \end{aligned}$
Thus our equation becomes
$\begin{aligned} &y e^{-x^{2} / 2}=\int-e^{t} d t+c\\ &y e^{-x^{2} / 2}=-e^{t}+c\\ &\text { Putting back } t=-\frac{x^{2}}{2}\\ &\Rightarrow \mathrm{ye}^{-\frac{\mathrm{x}^{2}}{2}}\\ &=-\mathrm{e}^{-\mathrm{x} / 2}+\mathrm{c}\\ &\Rightarrow \frac{y}{\mathrm{e}^{\mathrm{x}^{2} / 2}} \end{aligned}$
$\begin{aligned} &=-\frac{1}{e^{x^{2} / 2}}+c \\ &\Rightarrow y=-1+c e^{x^{2 / 2}}-(1) \end{aligned}$
Since curve passes through (0,1)
Putting
$\begin{aligned} &\mathrm{x}=0, \\ \\&\mathrm{y}=1 \mathrm{in}(1) \\ \\&\mathrm{y}=-1+\mathrm{ce}^{\mathrm{x}^{2} / 2} \\ \\&1=-1+\mathrm{ce}^{\mathrm{a}_{2} / 2} \\ \\&\Rightarrow 1+1=\mathrm{C} \end{aligned}$ $[\because e^{0}=1]$
$\Rightarrow c=2$
Putting value of c in (1)
$\begin{aligned} &y=-1+c e^{x^{2 / 2}} \\ &y=-1+2 e^{x^{2 / 2}} \end{aligned}$
$\therefore$ Equation of the curve is $-1+2e^{x^{2}/2}$

Differential Equation Exercise Revision Exercise (RE) Question 74

Answer : $x+y+1=e^{x}$
Given: The slope of the tangent to the curve at any point $(x , y)$ is equal to the sum of x coordinate and the product of x and y coordinate of that point
Hint: You must know about integrating factor
Explanation: We know that
Slope of the tangent to the curve at $(x , y)=\frac{dy}{dx}$
According to the given
$\begin{aligned} &\frac{d y}{d x}=x+x y \\ \\&\Rightarrow \frac{d y}{d x}-x y=x \end{aligned}$
This is of the form
$\begin{aligned} &\frac{d y}{d x}+Py=Q \\ \end{aligned}$
Where
$\begin{aligned} &\mathrm{P}=-\mathrm{1} \text { and }\\ \\&\theta=x\\ \end{aligned}$
Finding integrating factor
If,
$\begin{aligned} &=e^{\int P d x} \\ \\&=e^{\int(-1) d x} \\ \\&=e^{-x d x} \end{aligned}$ $\left [ \because \int 1\; dx =x+c\right ]$
Solution is
$\begin{aligned} &\mathrm{y} \cdot(\mathrm{I} \cdot \mathrm{f} \cdot)=\int(\theta \times \mathrm{I} \cdot \mathrm{f} \cdot) \mathrm{d} \mathrm{x}+\mathrm{c} \\ &\mathrm{ye}^{-\mathrm{x}}=\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \end{aligned}$
Using integration by parts,
$\begin{aligned} &\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}=\mathrm{x} \cdot \frac{\mathrm{e}^{-\mathrm{x}}}{-1}-\int(1) \frac{\mathrm{e}^{-\mathrm{x}}}{-1} \mathrm{dx} \\ &=-\mathrm{xe}^{-\mathrm{x}}+\frac{\mathrm{e}^{-\mathrm{x}}}{-1}+\mathrm{c} \\ &=-\mathrm{xe}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{C} \end{aligned}$
Put in (1)
$\Rightarrow y e^{-x}=-x e^{-x}-e^{-x}+c$
Divide by $e^{-x}$
$\Rightarrow y=-x-1+c e^{x}-(2)$
Since curve passes through origin,
Putting x=0 and y=0 in (2)
$\begin{gathered} 0 \quad=0-1+\mathrm{Ce}^{0} \\ \Rightarrow \mathrm{c}=1 \end{gathered}$ $[\because e^{0}=1]$
Put vaue of c in (2)
$\begin{aligned} &y=-x+1+e^{x} \\ \\&\Rightarrow x+y+1=e^{x} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 75

Answer: $y= \begin{cases}4-x-2 e^{x} & \text { iff } \frac{d y}{d x}=+(x+y-5) \\ 6-x-4 e^{-x} & \text { iff } \frac{d y}{d x}=-(x+4-5)\end{cases}$
Given: Sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Hint: Using integration by parts and integration factor
Explanation: We know that
Slope of the tangent to the curve at $(x , y)=\frac{dy}{dx}$
According to the given
$\begin{aligned} &\left|\frac{d y}{d x}\right|+5=x+y \\ &\Rightarrow\left|\frac{d y}{d x}\right|=x+y-5 \\ &\Rightarrow \frac{d y}{d x}=\pm(x+y-5) \end{aligned}$
So we will take both (+) ve and (-) ve sign and then solve it
Taking (+) ve sign:
$\begin{aligned} &\frac{d y}{d x}=x+y-5 \\ &\Rightarrow \frac{d y}{d x}-y=x-5 \end{aligned}$
Equation is of the form
$\frac{d y}{d x}+P_{y}=Q$
Where,
$\begin{aligned} &P=-1 \\ &\theta=x-5 \\ &I \cdot F \cdot=e^{\int P d x} \\ &\qquad=e^{\int(-1) d x} \\ &\; \; \; \; \quad=e^{-x} \end{aligned}$
Solution is
$\begin{aligned} &\mathrm{y}(\mathrm{I} \cdot \mathrm{F})=\int(\theta \times \mathrm{I} \cdot \mathrm{f} \cdot) \mathrm{d} \mathrm{x}+\mathrm{c} \\ &\mathrm{ye}^{-\mathrm{x}}=\int(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\int \quad(1) \frac{\mathrm{e}^{-\mathrm{x}}}{-1} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}+\int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}+\mathrm{c} \\ &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c} \end{aligned}$
Divided by $e^{-x}$
$\begin{aligned} &\Rightarrow \mathrm{y}=-(\mathrm{x}-5)-1+\mathrm{ce}^{\mathrm{x}}-(1) \\ \\&\Rightarrow \mathrm{y}_{\mathrm{y}}=5-\mathrm{x}-1+\mathrm{ce}^{\mathrm{x}} \\ \\&\; \; \; \; \; \; \quad=4-\mathrm{x}+\mathrm{ce}^{\mathrm{x}} \end{aligned}$
Since the curve passes through -the point (0,2)
Put x=0 and
$y=2$
$\begin{aligned} &2=4-0+c e^{\circ} \\ \\&\Rightarrow 2=4+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \mathbf{e}^{0}=1\right] \\ \\&\Rightarrow C=-2 \end{aligned}$
Equation of the curve is
$y=4-x-2 e^{x}$
Taking (-) ve sign:
$\begin{aligned} &\frac{d y}{d x}=-x-y+5 \\ &\Rightarrow \frac{d y}{d x}+y=-x+5 \end{aligned}$
Equation is of the form
$\frac{d y}{d x}+P_{y}=Q$
Where,
$\begin{aligned} &\mathrm{P}=1 \\ &\theta=-\mathrm{x}+5 \\ &\mathrm{I} \cdot \mathrm{F} \cdot=\mathrm{e}^{\int \mathrm{Pdx}} \\ &\qquad \begin{aligned} &=\mathrm{e}^{\int(1) \mathrm{dx}} \\ &=\mathrm{e}^{\mathrm{x}} \end{aligned} \end{aligned}$ $[\because \int 1dx=x+c]$
Solution is
$\begin{aligned} &y(I \cdot F)=\int(\theta \times I \cdot f \cdot) d x+c \\ &\Rightarrow y e^{x}=(5-x) e^{-x}-\int(-1) e^{x} d x+c \end{aligned}$ [Using integration by parts]
$\Rightarrow y e^{x}=(5-x) e^{-x}+e^{x}+c$
Divided by $e^{-x}$
$\begin{aligned} &\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}+\mathrm{c} \\ &\Rightarrow \mathrm{y}=-(\mathrm{x}-5)-1+\mathrm{ce}^{\mathrm{x}}-(1) \\ &\Rightarrow \mathrm{y}_{\mathrm{y}}=5-\mathrm{x}-1+\mathrm{ce}^{\mathrm{x}} \\ &\quad=4-\mathrm{x}+\mathrm{ce}^{\mathrm{x}} \end{aligned}$
Since the curve passes through the point (0,2)
Put x=0 and
$\begin{aligned} &y=2 \\ &2=6-0+c e^{\circ} \\ &\Rightarrow 2=6+c(1) \\ \end{aligned}$ $[\therefore e^{0}=1]$
$\Rightarrow C=-4$
Put in (2) we get,
Equation of the curve is
$y=6-x-4 e^{x}$

Differential Equation Exercise Revision Exercise (RE) Question 76

Answer : $y^{2}=x+5$
Given: The slope of the tangent to the curve at any point is the reciprocal of twice the ordinate at that point. Also the curve passes through the point(4,3)
Hint: Using variable separable method
Explanation: Let $P (x , y)$ be any point on the curve
Slop of the tangent at $P (x , y)=\frac{dy}{dx}$
$\therefore$ Acc to given condition,
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \\ \\&\Rightarrow 2 \mathrm{ydy}=1 \mathrm{dx} \end{aligned}$
Integrating both sides,
$\begin{aligned} &2 \mathrm{ydy}=\int 1 \mathrm{dx} \\ \\&\Rightarrow \frac{2 \mathrm{y}^{2}}{2}=\mathrm{x}+\mathrm{c} \\ \\&\Rightarrow \mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \end{aligned}$
Since the curve passes through (4,3) then,
$\\a=4+c\\\Rightarrow C=5$
Hence the required equation of the curve is
$y^{2}=x+5$

Differential Equation Exercise Revision Exercise (RE) Question 77

Answer : $\frac{\log 2}{\lambda}$
Given: The decay rate of radium at any time t is proportional to its mass at that time.
Hint: Using variable separable method
Explanation: Acc to given,
$\begin{aligned} &\frac{d m}{d t} \alpha m \\ &\Rightarrow \frac{d m}{d t}=-d m \end{aligned}$
$\begin{aligned} &\text { where } \lambda>0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { [-ve sign } \therefore \text { of decay rate }] \\ &\Rightarrow \frac{d m}{m}=-d t \end{aligned}$
Integrating both sides
$\begin{aligned} &\int \frac{\mathrm{dm}}{\mathrm{m}}=-\int \lambda \mathrm{dt}\\ &\Rightarrow \log \mathrm{m}=-\lambda+\mathrm{c}-(1)\\ &\text { Let } \mathrm{m}_{0} \text { be the initial mass at } \mathrm{t}=0\\ &\log m_{0}=-\lambda \cdot 0+c\\ &\Rightarrow \mathrm{c}=\log \mathrm{m}_{0} \end{aligned}$
Put in (1)
$\begin{aligned} &\Rightarrow \log \mathrm{m}=-\lambda \mathrm{t}+\log \mathrm{m}_{0} \\ &\Rightarrow \log \mathrm{m}-\log \mathrm{m}_{0} \\ &=-\lambda \mathrm{t} \\ &\Rightarrow \log \left(\frac{\mathrm{m}}{\mathrm{m}_{0}}\right)=-\lambda \mathrm{t}-2 \end{aligned}$ $\left[\because \log \mathbf{m}-\log \mathbf{x}=\log \mathbf{m}_{\mathrm{h}}\right]$
We have to find the time when $m=\frac{1}{2}m_{0}$
$\Rightarrow 2m=m_{0}$
Put in (2)
$\begin{aligned} &\Rightarrow \log \frac{\mathrm{m}}{2 \mathrm{~m}}=-\lambda \mathrm{t} \\ &\Rightarrow \log \frac{1}{2}=-\lambda \mathrm{t} \\ &\Rightarrow \log 1-\log 2=-\lambda \mathrm{t} \\ &\Rightarrow 0-\log 2=-\lambda \mathrm{t} \end{aligned}$ $[\because \log 1=0]$
$\begin{aligned} &\Rightarrow \lambda t=\log 2 \\ &\Rightarrow \mathrm{t}=\frac{\log 2}{\lambda} \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 78

Answer: 0.04%
Given: Radius disintegrates at a rate proportional to the amount of radium present at the moment
Hint: Using variable separable method
Explanation: Let A be the amount radium at that time
According to given,
$\begin{aligned} &\Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}} \propto \mathrm{A} \\ &\Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=-\lambda \mathrm{A}, \lambda>0 \\ &\Rightarrow \frac{\mathrm{dA}}{\mathrm{A}}=-\lambda \mathrm{dt} \end{aligned}$
Integrating both sides,
$\begin{aligned} &\int \frac{\mathrm{d} \mathrm{A}}{\mathrm{A}}=-\lambda \int \mathrm{dt} \\ &\Rightarrow \log _{\mathrm{e}} \mathrm{A}=-\lambda \mathrm{t}+\mathrm{c}-(1) \end{aligned}$
Let $A_{0}$ be the initial amount at $t=0$
$\begin{aligned} &\Rightarrow \log _{\mathrm{e}} \mathrm{A}_{0}=\mathrm{C} \\ &\therefore \log _{\mathrm{e}} \mathrm{A}=-\lambda \mathrm{t}+\log _{\mathrm{e}} \mathrm{A}_{0} \\ &\Rightarrow \log _{\mathrm{e}} \mathrm{A}-\log _{\mathrm{e}} \mathrm{A}_{0}=-\lambda \mathrm{t} \\ &\Rightarrow \log \mathrm{e} \frac{\mathrm{A}}{\mathrm{A}_{0}}=-\lambda \mathrm{t} \Rightarrow \log \frac{\mathrm{A}_{0}}{\mathrm{~A}}=\lambda \mathrm{t} \end{aligned}$
Now,
$\begin{aligned} &\mathrm{A}=\frac{\mathrm{A}_{\mathrm{o}}}{2} \\ \\&\mathrm{t}=1590 \\ \\&\Rightarrow \lambda \times 1590 \\ \\&=\log \frac{2 \mathrm{~A}}{\mathrm{~A}} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{\log 2}{1590}=\lambda \\ \\&\Rightarrow \lambda=\log _{c}\left(\frac{1+0}{A}\right) \\ \\&t=1 \\ \\&\Rightarrow \frac{\log 2}{1590}=\log c \end{aligned}$
$\begin{aligned} &\frac{A_{0}}{A} \Rightarrow \frac{A_{0}}{A}=e^{\frac{\log _{2}}{1590}} \\ \\&\Rightarrow A=e^{-\frac{l o g 2}{1590}} A_{0} \\ \\&\Rightarrow A=0.9996 A_{0} \end{aligned}$ $[\because \text {of given}]$
$\begin{aligned} \therefore \text { Reqd. } \%=& \frac{\mathrm{A}-\mathrm{A}_{0}}{\mathrm{~A}_{0}} \times 100 \\ =& \frac{0.9996 \mathrm{~A}_{0}-\mathrm{A}_{0}}{\mathrm{~A}_{0}} \times 100 \\ &=-0.04 \% \end{aligned}$

Differential Equation Exercise Revision Exercise (RE) Question 79

Answer : $\mathrm{T}=\frac{\log 20}{\log 2}$
Given: A wet porous substance in the wet air loses its moisture at a rate proportional to the moisture content
Hint: Using variable separable method
Explanation: Let M be the moisture content at any time A
According to given, $\begin{aligned} &\frac{\mathrm{d} \mathrm{M}}{\mathrm{dt}} \alpha \mathrm{M} \\ &\Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km} \end{aligned}$

when k be any constant and [-ve sign. ? of it losses its moisture]

$\Rightarrow \frac{d m}{m}=-k d t$

Integrating both the sides.

$\begin{aligned} &\int \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{k} \int \mathrm{dt}\\ &\log \mathrm{m}=-\mathrm{kt}+\mathrm{c}-(1)\\ &\text { Let at } \mathrm{t}=0 \text {, }\\ &\mathrm{M}=\mathrm{M}_{0} \end{aligned}$

Put in (1)

$\begin{aligned} &\log \mathrm{M}=-\mathrm{kt}+\log \mathrm{M}_{0} \\ &\Rightarrow \log \mathrm{M}-\log \mathrm{M}_{0}=-\mathrm{Kt} \\ &\Rightarrow \log \left|\frac{\mathrm{M}}{\mathrm{M}_{0}}\right|=-\operatorname{lct}-(2) \end{aligned}$ $\left[\because \log \mathrm{a}-\log \mathrm{b}=\frac{\log \mathrm{a}}{\mathrm{b}}\right]$

Given tht at $t=1,$

$\mathrm{m}=\frac{1}{2} \times \mathrm{m}_{0}$

Put in (2)

$\begin{aligned} &\Rightarrow \log \left|\frac{\mathrm{M}}{2 \mathrm{M}}\right|=-\mathrm{k} \times 1 \\ &\Rightarrow \log \frac{1}{2}=-\mathrm{K} \\ &\Rightarrow \log 2^{-1}=-\mathrm{k} \\ &\Rightarrow \log 2^{-1}=-\mathrm{k} \end{aligned}$ $\left[\because \log \mathbf{a}^{\mathbf{b}}=\text { blo g } \mathbf{a}\right]$

$\begin{aligned} &\Rightarrow-1 \cdot \log 2=-\mathrm{k} \\ &\mathrm{k}=\log 2 \end{aligned}$

Put in (2)

$\begin{aligned} &\log \left|\frac{\mathrm{M}}{\mathrm{M}_{0}}\right|=-\log 2 \cdot \mathrm{t} \\ &\Rightarrow \log \left|\frac{\mathrm{M}_{0}}{\mathrm{M}}\right|=\log 2 \cdot \mathrm{t}-(3) \end{aligned}$

Let at $t=T,$

moisture = 95%

$\Rightarrow$ Remaining = 5 %

$\begin{aligned} &\mathrm{M}=5 \% \text { of } \mathrm{M}_{0} \\ &\Rightarrow \mathrm{M}=\frac{5}{100} \times \mathrm{M}_{0} \\ &=\frac{1}{20} \mathrm{M}_{0} \end{aligned}$

Put in (3)

$\begin{aligned} &\log \left(\frac{20 \mathrm{M}}{\mathrm{M}}\right)=+\log 2 \cdot \mathrm{T} \\ &\Rightarrow \log 2_{0}=\log 2 \cdot \mathrm{T} \\ &\Rightarrow \mathrm{T}=\frac{\log 20}{\log 2} \end{aligned}$

Class 12 maths book consists of eleven exercises from the 21st chapter. There are 113 revision-based exercises in this material. The concepts on which these questions are framed are, differentiating various sums, integration factor of differential equations, formation of differential equations, solving first-order differential equations, and many more. It may seem to be quite enormous, but all it takes is a good practice with the RD Sharma Class 12th Exercise solution book. Therefore, the importance of the RD Sharma Class 12 Solutions Differential Equations Ex RE book is ineffable.

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