NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations: Exercise Questions

Here are the NCERT Class 12 Maths Chapter 9 Differential Equations question answers with clear and detailed solutions.

Question: Determine the order and degree (if defined) of the differential equation

1.$\frac{\mathrm{d}^4 y}{\mathrm{~d} x^4}+\sin \left(y^{\prime \prime \prime}\right)=0$

2. $y' + 5y = 0$

3.$\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$

4. $\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$

5. $\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$

6. $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$

7. $y''' + 2y'' + y' =0$

8. $y' + y = e^x$

9. $y'' + (y')^2 + 2y = 0$

10. $y'' + 2y' + \sin y = 0$

Answer(1):

The given function is
$\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$
We can rewrite it as
$y^{''''}+\sin(y''') =0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''''}$

Therefore, the order of the given differential equation $\frac{\mathrm{d} ^4y}{\mathrm{d} x^4} +\sin(y''')=0$ is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, its degree is not defined

Answer(2):

Given function is
$y' + 5y = 0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{'}$
Therefore, the order of the given differential equation $y' + 5y = 0$ is 1
Now, the given differential equation is a polynomial equation in its derivatives, and its highest power raised to y ' is 1
Therefore, its degree is 1.

Answer(3):

Given function is
$\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$
We can rewrite it as
$(s^{'})^4+3s.s^{''} =0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $s^{''}$

Therefore, the order of the given differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives, and the power raised to s '' is 1
Therefore, its degree is 1

Answer(4):

Given function is
$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$
We can rewrite it as
$(y^{''})^2+\cos y^{''} =0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}$

Therefore, the order of the given differential equation $\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$ is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, its degree is not defined

Answer(5):

Given function is
$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$
$\Rightarrow \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}\left ( \frac{d^2y}{dx^2} \right )$

Therefore, order of given differential equation $\frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives,
$ \frac {d^2y}{dx^2}$ and power raised to $\frac{d^2y}{dx^2}$ is 1
Therefore, its degree is 1.

Answer(6):

$
\text{Given function is}
\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^5=0
$

Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{\prime \prime \prime}.$

Therefore, the order of the given differential equation
$\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^5=0
\text{ is } 3.
$

Now, the given differential equation is a polynomial equation in its derivatives
$y^{\prime \prime \prime}, y^{\prime \prime}, y^{\prime},
\text{ and the power raised to } y^{\prime \prime \prime} \text{ is } 2.
$

$
\text{Therefore, its degree is } 2.
$

Answer(7):

$
\text{Given function is }
y''' + 2y'' + y' = 0
$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{'''}.
$
Therefore, the order of the given differential equation
$y''' + 2y'' + y' = 0 \text{ is } 3.
$

Now, the given differential equation is a polynomial equation in its derivatives
$y^{'''}, y^{''}, \text{ and } y^{'},
\text{ and the power raised to } y^{'''} \text{ is } 1.
$

$
\text{Therefore, its degree is } 1.
$

Answer(8):

Given function is
$y' + y = e^x$
$\Rightarrow$ $y^{'}+y-e^x=0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{'}$

Therefore, order of given differential equation $y^{'}+y-e^x=0$ is 1
Now, the given differential equation is a polynomial equation in its derivatives, and the power raised to $y^{'}$ is 1
Therefore, the issue is 1

Answer(9):

Given function is
$y'' + (y')^2 + 2y = 0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + (y')^2 + 2y = 0$ is 2
$
\text{Now, the given differential equation is a polynomial equation in its derivatives } y^{''} \text{ and } y^{'} \text{, and the power raised to } y^{''} \text{ is } 1.
$
Therefore, its degree is 1

Answer(10):

Given function is
$y'' + 2y' + \sin y = 0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + 2y' + \sin y = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives
$y^{''} \text{ and } y^{'} \text{, and the power raised to } y^{''} \text{ is } 1.$
Therefore, its degree is 1.

Question 11: The degree of the differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is

(A) 3

(B) 2

(C) 1

(D) not defined

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$
We can rewrite it as
$(y^{''})^3+(y^{'})^2+\sin y^{'}+1=0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}$

Therefore, the order of the given differential equation
$\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is 2.
Now, the given differential equation is not a polynomial derivative
Therefore, its degree is not defined.

Therefore, the answer is (D)

Question 12: The order of the differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is

(A) 2

(B) 1

(C) 0

(D) Not Defined

Answer:

Given function is
$2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$
We can rewrite it as
$2x.y^{''}-3y^{'}+y=0$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y^{''}$

Therefore, the order of the given differential equation
$2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is 2.

Therefore, the answer is (A)

Question: Verify that the given functions (explicit or implicit) are a solution of the corresponding differential equation:

1. $y = e^x + 1 \qquad :\ y'' -y'=0$

2. $y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0$

3. $y = \cos x + C\qquad :\ y' + \sin x = 0$

4. $y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}$

5. $y = Ax\qquad :\ xy' = y\;(x\neq 0)$

6. $y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)$

7. $xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)$

8. $y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y$

9. $x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0$

10. $y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)$

Answer(1):

Given,

$y = e^x + 1$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x$

Again, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y' }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x$

$\implies y'' = e^x$

Substituting the values of y’ and y'' in the given differential equations,

$y'' - y' = e^x - e^x = 0 =$ RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Answer(2):

Given,

$y = x^2 + 2x + C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 2x + C) = 2x + 2$

Substituting the values of y’ in the given differential equations,

$y' -2x - 2 =2x + 2 - 2x - 2 = 0= RHS$ .

Therefore, the given function is the solution of the corresponding differential equation.

Answer(3):

Given,

$y = \cos x + C$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cost + C) = -sinx$

Substituting the values of y’ in the given differential equations,

$y' - \sin x = -sinx -sinx = -2sinx \neq RHS$ .

Therefore, the given function is not the solution of the corresponding differential equation.

Answer(4):

Given,

$y = \sqrt{1 + x^2}$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{1 + x^2}) = \frac{2x}{2\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}$

Substituting the values of y in the RHS.

$\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Answer(5):

Given,

$y = Ax$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(Ax) = A$

Substituting the values of y' in LHS,

$xy' = x(A) = Ax = y = RHS$ .

Therefore, the given function is a solution of the corresponding differential equation.

Answer(6):

Given,

$y = x\sin x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x \sin x) = \sin x + x \cos x$

Substituting the values of y' in LHS,

$xy' = x(x \sin x+ x \cos x)$

Substituting the values of y in RHS.

$\\x \sin x + x\sqrt{x^2 - x^2 \sin^2 x} = x \sin x+ x^2\sqrt{1- \sin x^2} = x(\sin x+x \cos x) = LHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(7):

Given,

$xy = \log y + C$

Now, differentiating both sides w.r.t. x,

$\\ y + x\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(logy) = \frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}$
$ \\ \implies y^2 + xyy' = y' $
$\\ \implies y^2 = y'(1-xy) $
$ \\ \implies y' = \frac{y^2}{1-xy}$

Substituting the values of y' in LHS,

$y' = \frac{y^2}{1-xy} = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(8):

Given,

$y - cos y = x$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1$

$\implies$ y' + siny.y' = 1

$\implies$ y'(1 + siny) = 1

$\implies y' = \frac{1}{1+siny}$

Substituting the values of y and y' in LHS,

$(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})$

$= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}$

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Answer(9):

Given,

$x + y = \tan^{-1}y$

Now, differentiating both sides w.r.t. x,

$\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ $
$\\ \implies1+y^2 = y'(1-(1+y^2)) = -y^2y'$
$ \\ \implies y' = -\frac{1+y^2}{y^2}$

Substituting the values of y' in LHS,

$y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Answer(10):

Given,

$y = \sqrt{a^2 - x^2}$

Now, differentiating both sides w.r.t. x,

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$

Substituting the values of y and y' in LHS,

$x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS$

Therefore, the given function is a solution of the corresponding differential equation.

Question 11: The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4

Answer:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

Question 12: The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0

Answer:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constant

Question 1: Find the general solution: $\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

Answer:

Given,

$\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

$\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2}$
$ \implies dy = (sec^2\frac{x}{2} - 1)dx$

$\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx $
$ \implies y = 2tan^{-1}\frac{x}{2} - x + C$

Question 2: Find the general solution: $\frac{dy}{dx} = \sqrt{4-y^2}\ (-2 < y < 2)$

Answer:

Given the question

$\frac{dy}{dx} = \sqrt{4-y^2}$

$\\ \implies \frac{dy}{\sqrt{4-y^2}} = dx $
$ \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx$

$\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\$

The required general solution:

$\\ \implies sin^{-1}\frac{y}{2} = x + C$

Question 3: Find the general solution: $\frac{dy}{dx} + y = 1 (y\neq 1)$

Answer:

Given the question

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

$\\ \implies \frac{dy}{dx} = 1- y $
$ \implies \int\frac{dy}{1-y} = \int dx$

$(\int\frac{dx}{x} = lnx)$

$\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\ \implies log k(1-y) = -x $
$ \implies 1- y = \frac{1}{k}e^{-x} \\$

The required general equation

$\implies y = 1 -\frac{1}{k}e^{-x}$

Question:4 Find the general solution: $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

Answer:

Given,

$\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

$\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx $
$ \implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx$

Now, let tany = t and tax = u

$sec^2 y dy = dt\ and\ sec^2 x dx = du$

$\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} $
$ \implies log t = -log u +logk $
$ \implies t = \frac{1}{ku} $
$ \implies tany = \frac{1}{ktanx}$

Question:5 Find the general solution:

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

Answer:

Given, the question

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

$\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx$

Let,

$\\ (e^x + e^{-x}) = t \\ \implies (e^x - e^{-x})dx = dt$

$\\ \implies \int dy = \int \frac{dt}{t} $
$\implies y = log t + C $
$ \implies y = log(e^x + e^{-x}) + C$

This is the general solution

Question 6: Find the general solution: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

Answer:

Given, the question

$\frac{dy}{dx} = (1+x^2)(1+y^2)$

$\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx$

$(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)$

$\\ \implies tan^{-1}y = x+\frac{x^3}{3} + C$

Question:7 Find the general solution: $y\log y dx - x dy = 0$

Answer:

Given,

$y\log y dx - x dy = 0$

$\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx $
$ \implies \log t = \log x + \log k $
$\implies t = kx \\ \implies \log y = kx$

This is the general solution

Question 8: Find the general solution: $x^5\frac{dy}{dx} = - y^5$

Answer:

Given, the question

$x^5\frac{dy}{dx} = - y^5$

$\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} $
$ \implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C $
$ \implies \frac{1}{y^4} + \frac{1}{x^4} = C$

This is the required general equation.

Question 9: Find the general solution: $\frac{dy}{dx} = \sin^{-1}x$

Answer:

Given, the question

$\frac{dy}{dx} = \sin^{-1}x$

$\implies \int dy = \int \sin^{-1}xdx$

Now,

$\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx$

Here, u = $\sin^{-1}x$ and v = 1

$\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx$

$\\ Let\ 1- x^2 = t $
$ \implies -2xdx = dt $
$\implies xdx = -dt/2$

$\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) $
$ \implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C $
$\implies y = x\sin^{-1}x + \sqrt{1-x^2} + C$

Question 10: Find the general solution $e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

Answer:

Given,

$e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

$\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy $
$ \implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx$

$\\ let\ tany = t \ and \ 1-e^x = u $
$\implies \sec^2 ydy = dt\ and \ -e^xdx = du$

$\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} $
$ \implies \log t = \log u + \log k $
$ \implies t = ku$
$\implies \tan y= k (1-e^x)$

Question 11: Find a particular solution satisfying the given condition:

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0$

Answer:

Given, the question

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x$

$\\ \implies \int dy = \int\frac{2x^2 + x}{(x^3 + x^2 + x + 1)}dx$

$(x^3 + x^2 + x + 1) = (x +1)(x^2+1)$

Now,

$\begin{aligned} & \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1} \\ & \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A x^2+A(B x+C)(x+1)}{(x+1)\left(x^2+1\right)} \\ & \Rightarrow 2 x^2+x=A x^2+A+B x+C x+C \\ & \Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+A+C\end{aligned}$

Now, comparing the coefficients.

A + B = 2; B + C = 1; A + C = 0

Solving these:

$\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{3}{2}, \mathrm{C}=-\frac{1}{2}$

Putting the values of A, B, and C:

$\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{1}{2} \frac{1}{(x+1)}+\frac{1}{2} \frac{3 x-1}{x^2+1}$

Therefore,

$\begin{aligned} & \Rightarrow \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^2+1} d x \\ & \Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1} \\ & \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \tan ^{-1} \mathrm{x} \\ & \text { let } \mathrm{x}^2+1=\mathrm{t}\end{aligned}$

$\begin{aligned} & \therefore \frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}=\frac{3}{4} \int \frac{\mathrm{dt}}{\mathrm{t}} \\ & \text { so, } \mathrm{I}=\frac{3}{4} \log \mathrm{t} \\ & \mathrm{I}=\frac{3}{4} \log \left(\mathrm{x}^2+1\right) \\ & \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \log \left(\mathrm{x}^2+1\right)-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}\end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{y}=\frac{1}{4}\left[2 \log (\mathrm{x}+1)+3 \log \left(\mathrm{x}^2+1\right)\right]-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c} \\ & \Rightarrow \mathrm{y}=\frac{1}{4}\left[\log (\mathrm{x}+1)^2+\log \left(\mathrm{x}^2+1\right)^3\right]-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}\end{aligned}$

Now, y= 1 when x = 0

$1=\frac{1}{4} \times 0-\frac{1}{2} \times 0+c$

c = 1

Putting the value of c, we get:

$y=\frac{1}{4}\left[\log \left\{(x+1)^2\left(x^2+1\right)\right\}\right]-\frac{1}{2} \tan ^{-1} x+1$

Question 12: Find a particular solution satisfying the given condition:

$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$

Answer:

Given, the question

$x(x^2 -1)\frac{dy}{dx} =1$

$\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}$

Let,

$\begin{aligned} & \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{c}{x-1} \\ & \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)}{x(x+1)(x-1)} \\ & \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{(A+B+C) x^2+(B-C) x-A}{x(x+1)(x-1)}\end{aligned}$

Now, comparing the values of A, B, and C

A + B + C = 0; B-C = 0; A = -1

Solving these:

$B=\frac{1}{2}$ and $C=\frac{1}{2}$

Now, putting the values of A, B, C

$\begin{aligned} & \Rightarrow \frac{1}{x(x+1)(x-1)}=-\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{x-1}\right) \\ & \Rightarrow \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int\left(\frac{1}{x+1}\right) d x+\frac{1}{2} \int\left(\frac{1}{x-1}\right) d x \\ & \Rightarrow y=-\log x+\frac{1}{2} \log (x+1)+\frac{1}{2} \log (x-1)+\log c \\ & \left.\Rightarrow y=\frac{1}{2} \log \left[\frac{c^2(x-1)(x+1)}{x^2}\right\}-\text { iii }\right)\end{aligned}$

Given, y =0 when x =2

$\begin{aligned} & 0=\frac{1}{2} \log \left[\frac{c^2(2-1)(2+1)}{4}\right\} \\ & \Rightarrow \log \frac{3 c^2}{4}=0 \\ & \Rightarrow 3 c^2=4\end{aligned}$

Therefore,

$\\ \implies y = \frac{1}{2}\log[\frac{4(x-1)(x+1)}{3x^2}]$

$\\ \implies y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]$

Question 13: Find a particular solution satisfying the given condition:

$\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0$

Answer:

Given,

$\cos\left(\frac{dy}{dx} \right ) = a$

$\\ \implies \frac{dy}{dx} = \cos^{-1}a $
$ \implies \int dy = \int\cos^{-1}a\ dx $
$ \implies y = x\cos^{-1}a + c$

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

$\implies y = x\cos^{-1}a + 1$

Question 14: Find a particular solution satisfying the given condition:

$\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0$

Answer:

Given,

$\frac{dy}{dx} = y\tan x$

$\\ \implies \int \frac{dy}{y} = \int \tan x\ dx $
$ \implies \log y = \log \sec x + \log k $
$ \implies y = k\sec x$

Now, y=1 when x =0

1 = ksec0

$\implies$ k = 1

Putting the value of k:

y = sec x

Question 15: Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y' = e^x \sin x $.

Answer:

We first find the general solution of the given differential equation

Given,

$y' = e^x\sin x$

$\\ \implies \int dy = \int e^x\sin xdx$

$\\ Let I = \int e^x\sin xdx $
$ \implies I = \sin x.e^x - \int(\cos x. e^x)dx $
$\implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] $
$ \implies 2I = e^x(\sin x - \cos x) $
$ \implies I = \frac{1}{2}e^x(\sin x - \cos x)$

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c$

Now, since the curve passes through (0,0)

y = 0 when x =0

$\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\ \implies c = \frac{1}{2}$

Putting the value of c, we get:

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} $
$ \implies 2y -1 = e^x(\sin x - \cos x)$

Question 16: For the differential equation $xy\frac{dy}{dx} = (x+2)(y+2)$, find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

$xy\frac{dy}{dx} = (x+2)(y+2)$

$\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx $
$ \implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx $
$ \implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx $
$ \implies y - 2\log (y+2) = x + 2\log x + C$

Now, since the curve passes through (1,-1)

y = -1 when x = 1

$\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C $
$ \implies -1 -0 = 1 + 0 +C \\ \implies C = -2$

Putting the value of C:

$\\ y - 2\log (y+2) = x + 2\log x + -2 $
$ \implies y -x + 2 = 2\log x(y+2)$

Question 17: Find the equation of a curve passing through the point $( 0,-2)$ given that at any point $(x,y)$ on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

$y\frac{dy}{dx} =x$

$\\ \implies \int ydy =\int dx $
$ \implies \frac{y^2}{2} = \frac{x^2}{2} + c$

Now, since the curve passes through (0,-2).

x =0 and y = -2

$\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2$

Putting the value of c, we get

$\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4$

Question 18: 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). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is $\frac{y+3}{x+4}$

According to the question,

$\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) $
$\implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} $
$ \implies \log (y+3) = 2\log (x+4) + \log k $
$ \implies (y+3) = k(x+4)^2$

Now, since the curve passes through (-2,1)

x = -2 , y =1

$\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1$

Putting the value of k, we get

$\\ \implies y+3 = (x+4)^2$

Question 19: The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Answer:

Volume of a sphere, $V = \frac{4}{3}\pi r ^3$

Given that the rate of change is constant.

$\\ \therefore \frac{dV}{dt} = c $
$ \implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c $
$ \implies \int d(\frac{4}{3}\pi r ^3) = c\int dt $
$\implies \frac{4}{3}\pi r ^3 = ct + k$

Now, at t=0, r=3 and at t=3 , r =6

Putting these values:

$\frac{4}{3}\pi (3) ^3 = c(0) + k $
$\implies k = 36\pi$

Also,

$\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi $
$ \implies 3c = 252\pi $
$ \implies c = 84\pi$

Putting the value of c and k:

$\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi $
$ \implies r ^3 = (21 t + 9)(3) = 62t + 27 $
$ \implies r = \sqrt[3]{62t + 27}$

Question 20: In a bank, the principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 doubles itself in 10 years (log e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to the question,

$\frac{dp}{dt} = (\frac{r}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt $
$ \implies \log p = \frac{r}{100}t + C$

$\\ \implies p = e^{\frac{rt}{100} + C}$

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

$\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C$

Also,

$\\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\ \implies e^{\frac{r}{10}} = 2$
$ \implies \frac{r}{10} = \ln 2 = 0.6931 $
$ \implies r = 6.93$

So value of r = 6.93%

Question 21: In a bank, the principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it be worth after 10 years (e 0.5 = 1.648)?

Answer:

Let p be the principal amount and t be the time.

According to the question,

$\frac{dp}{dt} = (\frac{5}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt$
$ \implies \log p = \frac{1}{20}t + C$

$\\ \implies p = e^{\frac{t}{20} + C}$

Now, at t =0 , p = 1000

Putting these values,

$\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C$

Also, at t=10

$\\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 $
$ \implies p =(1.648)(1000) = 1648$

After 10 years, the total amount would be Rs. 1648.

Question 22: In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to the question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt $
$ \implies \log n = kt + C$

Now, at t=0, n = 100000

$\\ \implies \log (100000) = k(0) + C \\ \implies C = 5$

Again, at t=2, n= 110000

$\\ \implies \log (110000) = k(2) + 5 $
$ \implies \log 11 + 4 = 2k + 5 $
$ \implies 2k = \log 11 -1 =\log \frac{11}{10} $
$ \implies k = \frac{1}{2}\log \frac{11}{10}$

Using these values, for n= 200000

$\\ \implies \log (200000) = kt + C $
$ \implies \log 2 +5 = kt + 5 $
$ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 $
$ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}$

Question 23: The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

(A) $e^x + e^{-y} = C$

(B) $e^{x }+ e^{y} = C$

(C) $e^{-x }+ e^{y} = C$

(D) $e^{-x }+ e^{-y} = C$

Answer:

Given,

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

$\\ \implies\frac{dy}{dx} = e^x.e^y $
$ \implies\int \frac{dy}{e^y} = \int e^x.dx $
$\implies -e^{-y} = e^x + C $
$\implies e^x + e^{-y} = K$ (Option A)

Question 1: Show that the given differential equation is homogeneous and solve each of them. $(x^2 + xy)dy = (x^2 + y^2)dx$

Answer:

The given differential equation can be written as
$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Let $F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Now, $F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}$
$=\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y)$ Hence, it is a homogeneous equation.

To solve it, put y = vx
differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}$

$x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}$

$( \frac{1+v}{1-v})dv = \frac{dx}{x}$

$( \frac{2}{1-v}-1)dv = \frac{dx}{x}$
Integrating on both sides, we get;
$\\-2\log(1-v)-v=\log x -\log k$
$ v= -2\log (1-v)-\log x+\log k$
$ v= \log\frac{k}{x(1-v)^{2}}\\$
Again substitute the value $y = \frac{v}{x}$ ,we get;

$\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}$
This is the required solution for the given diff. equation

Question 2: Show that the given differential equation is homogeneous and solve each of them. $y' = \frac{x+y}{x}$

Answer:

The above differential equation can be written as,

$\frac{dy}{dx} = F(x,y)=\frac{x+y}{x}$ ............................(i)

Now, $F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)$
Thus, the given differential equation is a homogeneous equation
Now, to solve, substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v$
$\\x\frac{dv}{dx}= 1\\ dv = \frac{dx}{x}$
Integrating on both sides, we get; (and substitute the value of $v =\frac{y}{x}$ )

$\\v =\log x+C\\ \frac{y}{x}=\log x+C\\ y = x\log x +Cx$
This is the required solution

Question 3: Show that the given differential equation is homogeneous and solve each of them.

$(x-y)dy - (x+y)dx = 0$

Answer:

The given differential eq can be written as;

$\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say)$ ....................................(i)

$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)$
Hence, it is a homogeneous equation.

Now, to solve, substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= \frac{1+v}{1-v}$
$ x\frac{dv}{dx} = \frac{1+v}{1-v}-v =\frac{1+v^{2}}{1-v}$

$\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}$
Integrating on both sides, we get;

$\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C$
again substitute the value of $v=y/x$
$\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C$
$ \tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C$
$ tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C$ This is the required solution.

Question 4: Show that the given differential equation is homogeneous and solve each of them.

$(x^2 - y^2)dx + 2xydy = 0$

Answer:

We can write it as;

$\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say)$ ...................................(i)

$F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)$
Hence, it is a homogeneous equation

Now, to solve the substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}$
$\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}$
integrating on both sides, we get

$\log (1+v^{2})= -\log x +\log C = \log C/x$
$\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx$ .............[ $v =y/x$ ]
This is the required solution.

Question 5: Show that the given differential equation is homogeneous and solve it.

$x^2\frac{dy}{dx} = x^2 - 2y^2 +xy$

Answer:

$\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)$

$F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y)$ ............(i)
Hence, it is a homogeneous equation

Now, to solve the substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= 1-2v^{2}+v$
$ x\frac{dv}{dx} = 1-2v^{2}$
$\frac{dv}{1-2v^{2}}=\frac{dx}{x}$

$1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}$

On integrating both sides, we get;

$\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C$
after substituting the value of $v= y/x$

$\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C$

This is the required solution

Question 6: Show that the given differential equation is homogeneous and solve it.

$xdy - yd= \sqrt{x^2 + y^2}dx$

Answer:

$\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y)$ .................................(i)

$F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)$
Hence is a homogeneous equation

Now, to solve, use substitution y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$
Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v+\sqrt{1+v^{2}}=\sqrt{1+v^{2}}$

$=\frac{dv}{\sqrt{1+v^{2}}} =\frac{dx}{x}$

On integrating both sides,

$\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C$
Substituting the value of v=y/x, we get

$\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\ y+\sqrt{x^{2}+y^{2}} = Cx^{2}$

Required solution

Question 7: Solve.

$\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy$

Answer:

$\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y)$ ......................(i)
By looking at the equation, we can directly say that it is a homogeneous equation.

Now, to solve, use substitution y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}$
$ =x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}$
$ =(\tan v-1/v)dv = \frac{2dx}{x}$

integrating on both sides, we get

$\\=\log(\frac{\sec v}{v})= \log (Cx^{2})$
$=\sec v/v =Cx^{2}$
substitute the value of v= y/x , we get

$\\\sec(y/x) =Cxy \\ xy \cos (y/x) = k$

Required solution

Question 8: Solve.

$x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0$

Answer:

$\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)$
It is a homogeneous equation

Now, to solve, use substitution y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v- \sin v = -\sin v$
$\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}$

On integrating both sides, we get;

$\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C$
$ \Rightarrow cosec (y/x) - \cot (y/x) = C/x$

$= x[1-\cos (y/x)] = C \sin (y/x)$ Required solution

Question 9: Solve.

$ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0$

Answer:

$\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y)$ ..................(i)

$\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)$

Hence, it is a homogeneous equation

Now, to solve, use substitution y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}$
$=x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}$
$ =[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}$
integrating on both sides, we get: ( substituting v =y/x)

$\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)$
$\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\$
$ \Rightarrow \log (y/x)-1=Cy$

This is the required solution of the given differential equation

Question 10: Solve.

$\left(1 + e^{\frac{x}{y}} \right )dx + e^\frac{x}{y}\left(1-\frac{x}{y}\right )dy = 0$

Answer:

$\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y)$ .......................................(i)

$= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)$
Hence, it is a homogeneous equation.

Now, to solve, use substitution x = yv

Differentiating on both sides wrt $x$
$\frac{dx}{dy}= v +y\frac{dv}{dy}$

Substitute this value in equation (i)

$\\=v+y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}} $
$ =y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}$
$ =\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}$

Integrating on both sides, we get;

${100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\ =[\frac{x}{y}+e^{x/y}]= \frac{c}{y}$
$\Rightarrow x+ye^{x/y}=c$
This is the required solution of the diff equation.

Question 11: Solve for a particular solution.

$(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1$

Answer:

$\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y)$ ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve, use substitution y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$
Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}$
$\Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}$

$\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}$

On integrating both sides

$⇒\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k$
$ ⇒\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k$
$⇒\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k$
$ ⇒\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k$ ......................(ii)

Now, y=1 and x= 1
$\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\$

After substituting the value of 2k in the equation. (ii)

$\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2$
This is the required solution.

Question 12: Solve for a particular solution.

$x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1$

Answer:

$\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)$
Hence, it is a homogeneous equation

Now, to solve, use substitution y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substituting this value in equation (i), we get

$⇒v+\frac{xdv}{dx}= -v- v^{2}$
$⇒\frac{xdv}{dx}=-v(v+2)$
$ ⇒\frac{dv}{v+2}=-\frac{dx}{x}$
$ ⇒\frac12[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}$

Integrating on both sides, we get;

$\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}$

Replace the value of v=y/x

$\frac{x^{2}y}{y+2x}=C^{2}$ .............................(ii)

Now y =1 and x = 1

$C = 1/\sqrt{3}$
therefore,

$\frac{x^{2}y}{y+2x}=1/3$

Required solution

Question 13: Solve for a particular solution.

$\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1$

Answer:

$\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y)$ ..................(i)

$F(\mu x,\mu y)=\frac{-[\mu x\sin^{2}(\mu y/\mu x)-\mu y]}{\mu x}=\mu ^{0}.F(x,y)$

Hence, it is a homogeneous equation

Now, to solve, use substitution y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

On integrating both sides, we get;

$\\-\cot v =\log\left | x \right | -C\\ =\cot v = \log\left | x \right |+\log C$

On substituting v =y/x

$=\cot (y/x) = \log\left | Cx \right |$ ............................(ii)

Now, $y = \pi/4\ @ x=1$

$\\\cot (\pi/4) = \log C \\ =C=e^{1}$

Put this value of C in Eq. (ii)

$\cot (y/x)=\log\left | ex \right |$

Required solution.

Question 14: Solve for a particular solution.

$\frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1$

Answer:

$\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y)$ ....................................(i)

The above equation is homogeneous. So,
Now, to solve, use substitution y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}$

On integrating both sides, we get;

$\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx$ .................................(ii)

now y = 0 and x =1 , we get

$C =e^{1}$

Put the value of C in Eq. 2

$\cos(y/x)=\log \left | ex \right |$

Question 15: Solve for a particular solution.

$2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1$

Answer:

The above equation can be written as:

$\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)$
By looking, we can say that it is a homogeneous equation.

Now, to solve, use substitution y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}$

integrating on both sides, we get;

$\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C$ .............................(ii)

Now, y = 2 and x =1, we get

C =-1
Put this value in equation(ii)

$\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}$

Question:16 A homogeneous differential equation of the from $\frac{dx}{dy}= h\left(\frac{x}{y} \right )$ can be solved by making the substitution.

(A) $y = vx$

(B) $v = yx$

(C) $x = vy$

(D) $x =v$

Answer:

$\frac{dx}{dy}= h\left(\frac{x}{y} \right )$
To solve this type of equation, put x/y = v
x = vy

Option C is correct

Question 17: Which of the following is a homogeneous differential equation?

(A) $(4x + 6x +5)dy - (3y + 2x +4)dx = 0$

(B) $(xy)dx - (x^3 + y^3)dy = 0$

(C) $(x^3 +2y^2)dx + 2xydy =0$

(D) $y^2dx + (x^2 -xy -y^2)dy = 0$

Answer:

Option D is the right answer.

$y^2dx + (x^2 -xy -y^2)dy = 0$
$\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)$
We can take out lambda as a common factor, and it can be cancelled out

Question:1 Find the general solution:

$\frac{dy}{dx} + 2y = \sin x$

Answer:

The given equation is
$\frac{dy}{dx} + 2y = \sin x$
This is $\frac{dy}{dx} + py = Q$ type where p = 2 and Q = sin x
Now,
$I.F. = e^{\int pdx}= e^{\int 2dx}= e^{2x}$
Now, the solution of a given differential equation is given by the relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{\int 2x }) =\int (\sin x\times e^{\int 2x })dx +C$
Let $I =\int (\sin x\times e^{\int 2x })$
$I = \sin x \int e^{2x}dx- \int \left ( \frac{d(\sin x)}{dx}.\int e^{2x}dx \right )dx$
$ I = \sin x.\frac{e^{2x}}{2}- \int \left ( \cos x.\frac{e^{2x}}{2} \right )$
$ I = \sin x. \frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x\int e^{2x}dx- \left ( \frac{d(\cos x)}{dx}.\int e^{2x}dx \right ) \right )dx$
$ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+ \int \left ( \sin x.\frac{e^{2x}}{2} \right ) \right )$
$ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+\frac{I}{2} \right ) $
$ (\because I = \int \sin xe^{2x})$
$\frac{5I}{4}= \frac{e^{2x}}{4}\left ( 2\sin x-\cos x \right )$
$ I = \frac{e^{2x}}{5}\left ( 2\sin x-\cos x \right )$
Put the value of I in our equation
Now, our equation becomes
$Y.e^{x^2 }= \frac{e^{2x}}{5}\left (2 \sin x-\cos x \right )+C$
$Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$
Therefore, the general solution is $Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}$

Question:2 Solve for general solution:

$\frac{dy}{dx} + 3y = e^{-2x}$

Answer:

The given equation is
$\frac{dy}{dx} + 3y = e^{-2x}$
This is $\frac{dy}{dx} + py = Q$ type where p = 3 and $Q = e^{-2x}$
Now,
$I.F. = e^{\int pdx}= e^{\int 3dx}= e^{3x}$
Now, the solution of the given differential equation is given by the relation
$Y(I.F.) =\int (Q\times I.F.)dx +C$
$Y(e^{ 3x }) =\int (e^{-2x}\times e^{ 3x })dx +C$
$Y(e^{ 3x }) =\int (e^{x})dx +C\\ Y(e^{3x})= e^x+C\\ Y = e^{-2x}+Ce^{-3x}$
Therefore, the general solution is $Y = e^{-2x}+Ce^{-3x}$

Question:3 Find the general solution

$\frac{dy}{dx} + \frac{y}{x} = x^2$

Answer:

The given equation is
$\frac{dy}{dx} + \frac{y}{x} = x^2$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x}$ and $Q = x^2$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x}dx}= e^{\log x}= x$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x) =\int (x^2\times x)dx +C$
$y(x) =\int (x^3)dx +C\\ y.x= \frac{x^4}{4}+C\\$
Therefore, the general solution is $yx =\frac{x^4}{4}+C$

Question 4: Solve for General Solution.

$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$

Answer:

The given equation is
$\frac{dy}{dx} + (\sec x)y = \tan x \ \left(0\leq x < \frac{\pi}{2} \right )$
This is $\frac{dy}{dx} + py = Q$ type where $p = \sec x$ and $Q = \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec xdx}= e^{\log |\sec x+ \tan x|}= \sec x+\tan x$
$(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec x+\tan x) =\int ((\sec x+\tan x)\times \tan x)dx +C$
$y(\sec x+ \tan x) =\int (\sec x\tan x+\tan^2 x)dx +C$
$y(\sec x+ \tan x) =\sec x+\int (\sec^2x-1)dx +C$
$ y(\sec x+ \tan x) = \sec x +\tan x - x+C$
Therefore, the general solution is $y(\sec x+ \tan x) = \sec x +\tan x - x+C$

Question:5 Find the general solution.

$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$

Answer:

The given equation is
$\cos^2 x\frac{dy}{dx} + y = \tan x\left(0\leq x < \frac{\pi}{2} \right )$
We can rewrite it as
$\frac{dy}{dx}+\sec^2x y= \sec^2x\tan x$
This is $\frac{dy}{dx} + py = Q$ where $p = \sec ^2x$ and $Q =\sec^2x \tan x$
Now,
$I.F. = e^{\int pdx}= e^{\int \sec^2 xdx}= e^{\tan x}$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(e^{\tan x}) =\int ((\sec^2 x\tan x)\times e^{\tan x})dx +C$
$ye^{\tan x} =\int \sec^2 x\tan xe^{\tan x}dx+C\\$
take
$e^{\tan x } = t\\ \Rightarrow \sec^2x.e^{\tan x}dx = dt$
$\int t.\log t dt = \log t.\int tdt-\int \left ( \frac{d(\log t)}{dt}.\int tdt \right )dt $
$ \int t.\log t dt = \log t . \frac{t^2}{2}- \int (\frac{1}{t}.\frac{t^2}{2})dt$
$ \int t.\log t dt = \log t.\frac{t^2}{2}- \int \frac{t}{2}dt$
$ \int t.\log t dt = \log t.\frac{t^2}{2}- \frac{t^2}{4}$
$ \int t.\log t dt = \frac{t^2}{4}(2\log t -1)$
Now put again $t = e^{\tan x}$
$\int \sec^2x\tan xe^{\tan x}dx = \frac{e^{2\tan x}}{4}(2\tan x-1)$
Put this value in our equation

$ye^{\tan x} =\frac{e^{2\tan x}}{4}(2\tan x-1)+C$
Therefore, the general solution is $y =\frac{e^{\tan x}}{4}(2\tan x-1)+Ce^{-\tan x }\\$

Question:6 Solve for General Solution.

$x\frac{dy}{dx} + 2y = x^2\log x$

Answer:

The given equation is
$x\frac{dy}{dx} + 2y = x^2\log x$
We can rewrite it as
$\frac{dy}{dx} +2.\frac{y}{x}= x\log x$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{2}{x}$ and $Q = x\log x$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2}{x}dx}= e^{2\log x}=e^{\log x^2} = x^2$
$(\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x^2) =\int (x\log x\times x^2)dx +C$
$x^2y = \int x^3\log x+ C$
Let
$I = \int x^3\log x$
$ I = \log x\int x^3dx-\int \left ( \frac{d(\log x)}{dx}.\int x^3dx \right )dx$
$ I = \log x.\frac{x^4}{4}- \int \left ( \frac{1}{x}.\frac{x^4}{4} \right )dx$
$I = \log x.\frac{x^4}{4}- \int \left ( \frac{x^3}{4} \right )dx$
$I = \log x.\frac{x^4}{4}-\frac{x^4}{16}$
Put this value in our equation
$x^2y =\log x.\frac{x^4}{4}-\frac{x^4}{16}+ C$
$y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$
Therefore, the general solution is $y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}$

Question 7: Solve for general solutions.

$x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$

Answer:

The given equation is
$x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$
We can rewrite it as
$\frac{dy}{dx}+\frac{y}{x\log x}= \frac{2}{x^2}$
This is $\frac{dy}{dx} + py = Q$ type where $p = \frac{1}{x\log x}$ and $Q =\frac{2}{x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{1}{x\log x} dx}= e^{\log(\log x)} = \log x$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\log x) =\int ((\frac{2}{x^2})\times \log x)dx +C$

take
$I=\int ((\frac{2}{x^2})\times \log x)dx$
$I = \log x.\int \frac{2}{x^2}dx-\int \left ( \frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx \right )dx $
$ I= -\log x . \frac{2}{x}+ \int (\frac{1}{x}.\frac{2}{x})dx$
$ I = -\log x.\frac{2}{x}+ \int \frac{2}{x^2}dx$
$I = -\log x.\frac{2}{x}- \frac{2}{x}\\ \\$
Put this value in our equation

$y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$
Therefore, the general solution is $y\log x =-\frac{2}{x}(\log x+1)+C\\ \\$

Question 8: Find the general solution.

$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$

Answer:

The given equation is
$(1 + x^2)dy + 2xydx = \cot x dx\ (x\neq 0)$
We can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}= \frac{\cot x}{1+x^2}$
This is $\frac{dy}{dx} + py = Q$ type
where $p = \frac{2x}{1+ x^2}$ and $Q =\frac{\cot x}{1+x^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+ x^2} dx}= e^{\log(1+ x^2)} = 1+x^2$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+x^2) =\int ((\frac{\cot x}{1+x^2})\times (1+ x^2))dx +C$
$y(1+x^2) =\int \cot x dx+C$
$ y(1+x^2)= \log |\sin x|+ C$
$ y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$
Therefore, the general solution is $y = (1+x^2)^{-1}\log |\sin x|+C(1+x^2)^{-1}$

Question 9: Solve for a general solution.

$x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)$

Answer:

The given equation is
$x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)$
We can rewrite it as
$\frac{dy}{dx}+y.\left ( \frac{1}{x}+\cot x \right )= 1$
This is $\frac{dy}{dx} + py = Q$ type
where $p =\left ( \frac{1}{x}+\cot x \right )$ and $Q =1$
Now,
$I.F. = e^{\int pdx}= e^{\int \left ( \frac{1}{x}+\cot x \right ) dx}= e^{\log x +\log |\sin x|} = x.\sin x$
Now, the solution of the given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(x.\sin x) =\int 1\times x\sin xdx +C$
$y(x.\sin x) =\int x\sin xdx +C$
Let's take
$I=\int x\sin xdx$
$ I = x .\int \sin xdx-\int \left ( \frac{d(x)}{dx}.\int \sin xdx \right )dx$
$ I =- x.\cos x+ \int (\cos x)dx$
$I = -x\cos x+\sin x$
Put this value in our equation
$y(x.\sin x)= -x\cos x+\sin x + C\\ y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$
Therefore, the general solution is $y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}$

Question 10: Find the general solution.

$(x+y)\frac{dy}{dx} = 1$

Answer:

The given equation is
$(x+y)\frac{dy}{dx} = 1$
We can rewrite it as
$\frac{dy}{dx} = \frac{1}{x+y}$
$x+ y =\frac{dx}{dy}$
$ \frac{dx}{dy}-x=y$
This is $\frac{dx}{dy} + px = Q$ type
where $p =-1$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int -1 dy}= e^{-y}$
Now, the solution of a given differential equation is given by the relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(e^{-y}) =\int y\times e^{-y}dy +C$
$xe^{-y}= \int y.e^{-y}dy + C$
Let's take
$I=\int ye^{-y}dy $
$ I = y .\int e^{-y}dy-\int \left ( \frac{d(y)}{dy}.\int e^{-y}dy \right )dy$
$I =- y.e^{-y}+ \int e^{-y}dy$
$ I = - ye^{-y}-e^{-y}$
Put this value in our equation
$x.e^{-y} = -e^{-y}(y+1)+C$
$ x = -(y+1)+Ce^{y}\\ x+y+1=Ce^y$
Therefore, the general solution is $x+y+1=Ce^y$

Question 11: Solve for a general solution.

$y dx + (x - y^2)dy = 0$

Answer:

The given equation is
$y dx + (x - y^2)dy = 0$
We can rewrite it as
$\frac{dx}{dy}+\frac{x}{y}=y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{1}{y}$ and $Q =y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{1}{y} dy}= e^{\log y } = y$
Now, the solution of a given differential equation is given by the relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(y) =\int y\times ydy +C$
$xy= \int y^2dy + C$
$xy = \frac{y^3}{3}+C$
$x = \frac{y^2}{3}+\frac{C}{y}$
Therefore, the general solution is $x = \frac{y^2}{3}+\frac{C}{y}$

Question 12: Find the general solution.

$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$

Answer:

The given equation is
$(x+3y^2)\frac{dy}{dx} = y\ (y > 0)$
We can rewrite it as
$\frac{dx}{dy}-\frac{x}{y}= 3y$
This is $\frac{dx}{dy} + px = Q$ type where $p =\frac{-1}{y}$ and $Q =3y$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{-1}{y} dy}= e^{-\log y } =y^{-1}= \frac{1}{y}$
Now, the solution of a given differential equation is given by the relation
$x(I.F.) =\int (Q\times I.F.)dy +C$
$x(\frac{1}{y}) =\int 3y\times \frac{1}{y}dy +C$
$\frac{x}{y}= \int 3dy + C$
$\frac{x}{y}= 3y+ C$
$x = 3y^2+Cy$
Therefore, the general solution is $x = 3y^2 + Cy$

Question 13: Solve for a particular solution.

$\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}$

Answer:

The given equation is
$\frac{dy}{dx} + 2y \tan x = \sin x; \ y = 0 \ when \ x =\frac{\pi}{3}$
This is $\frac{dy}{dx} + py = Q$ type
where $p = 2\tan x$ and $Q = \sin x$
Now,
$I.F. = e^{\int pdx}= e^{\int 2\tan xdx}= e^{2\log |\sec x|}= \sec^2 x$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sec^2 x) =\int ((\sin x)\times \sec^2 x)dx +C$
$y(\sec^2 x) =\int (\sin \times \frac{1}{\cos x}\times \sec x)dx +C$
$y(\sec^2 x) = \int \tan x\sec xdx+ C$
$ y.\sec^2 x= \sec x+C$
Now, by using boundary conditions, we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{3}$
at $x= \frac{\pi}{3}$
$0.\sec \frac{\pi}{3} = \sec \frac{\pi}{3}+C$
$ C = - 2$
Now,

$y.\sec^2 x= \sec x - 2\\ \frac{y}{\cos ^2x}= \frac{1}{\cos x}- 2\\ y = \cos x- 2\cos ^2 x$
Therefore, the particular solution is $y = \cos x- 2\cos ^2 x$

Question 14: Solve for a particular solution.

$(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1$

Answer:

The given equation is
$(1 + x^2)\frac{dy}{dx} + 2xy =\frac{1}{1 + x^2}; \ y = 0 \ when \ x = 1$
We can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2)^2}$
This is $\frac{dy}{dx} + py = Q$ type
where $p =\frac{2x}{1+x^2}$
and $Q = \frac{1}{(1+x^2)^2}$
Now,
$I.F. = e^{\int pdx}= e^{\int \frac{2x}{1+x^2}dx}= e^{\log |1+x^2|}= 1+x^2$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(1+ x^2) =\int (\frac{1}{(1+x^2)^2}\times (1+x^2))dx +C$
$y(1+x^2) =\int \frac{1}{(1+x^2)}dx +C$
$ \\ y(1+x^2) = \tan^{-1}x+ C\\ \\$
Now, by using boundary conditions, we will find the value of C
It is given that y = 0 when x = 1
at x = 1
$0.(1+1^2) = \tan^{-1}1+ C$
$C =- \frac{\pi}{4}$
Now,
$y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$
Therefore, the particular solution is $y(1+x^2)= \tan^{-1}x- \frac{\pi}{4}$

Question 15: Find the particular solution.

$\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}$

Answer:

The given equation is
$\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}$
This is $\frac{dy}{dx} + py = Q$ type
where $p =-3\cot x$ and $Q =\sin 2x$
Now,
$I.F. = e^{\int pdx}= e^{-3\cot xdx}= e^{-3\log|\sin x|}= \sin ^{-3}x= \frac{1}{\sin^3x}$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\frac{1}{\sin^3 x}) =\int (\sin 2x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\sin x\cos x\times\frac{1}{\sin^3 x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times \frac{\cos x}{\sin x}\times\frac{1}{\sin x})dx +C$
$\frac{y}{\sin^3 x} =\int (2\times\cot x\times cosec x)dx +C$
$\frac{y}{\sin^3 x} =-2cosec x +C$
Now, by using boundary conditions, we will find the value of C
It is given that y = 2 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$\frac{2}{\sin^3\frac{\pi}{2}} = -2cosec \frac{\pi}{2}+C$
$ 2 = -2 +C\\ C = 4$
Now,
$y= 4\sin^3 x-2\sin^2x\\$
Therefore, the particular solution is $ y=4\sin^3 x-2\sin^2x$

Question 16: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let f(x, y) be the curve passing through the origin
Then, the slope of the tangent to the curve at the point (x, y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} = y + x$
$ \\ \frac{dy}{dx}-y=x$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int (x \times e^{-x} )dx+ C$
Now, Let
$I= \int (x \times e^{-x} )dx $
$ \\ I = x.\int e^{-x}dx-\int \left ( \frac{d(x)}{dx}.\int e^{-x}dx \right )dx$
$ \\ I = -xe^{-x}+\int e^{-x}dx\\ $
$\\ I = -xe^{-x}-e^{-x}$
$ \\ I = -e^{-x}(x+1)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x+1)+C$
Now, by using boundary conditions, we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
$0.e^{-0}= -e^{-0}(0+1)+C$
$ C = 1$
Our final equation becomes
$ye^{-x}= -e^{-x}(x+1)+1$
$ y+x+1=e^x$
Therefore, the required equation of the curve is $y+x+1=e^x$

Question 17: Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let f(x, y) be the curve passing through the point (0, 2)
Then, the slope of the tangent to the curve at the point (x, y) is given by $\frac{dy}{dx}$
Now, it is given that
$\frac{dy}{dx} +5= y + x$
$ \\ \frac{dy}{dx}-y=x-5$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = -1 \ and \ Q = x- 5$
Now,
$I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}$
Now,
$y(I.F.)= \int (Q \times I.F. )dx+ C$
$y(e^{-x})= \int ((x-5) \times e^{-x} )dx+ C$
Now, Let
$I= \int ((x-5) \times e^{-x} )dx$
$ I = (x-5).\int e^{-x}dx-\int \left ( \frac{d(x-5)}{dx}.\int e^{-x}dx \right )dx$
$ I = -(x-5)e^{-x}+\int e^{-x}dx$
$ I = -xe^{-x}-e^{-x}+5e^{-x}$
$ I = -e^{-x}(x-4)$
Put this value in our equation
$ye^{-x}= -e^{-x}(x-4)+C$
Now, by using boundary conditions, we will find the value of C
It is given that the curve passes through the point (0, 2)
$2.e^{-0}= -e^{-0}(0-4)+C$
$C = -2$
Our final equation becomes
$ye^{-x}= -e^{-x}(x-4)-2$
$ y=4-x-2e^x$
Therefore, the required equation of the curve is $y=4-x-2e^x$

Question 18: The Integrating Factor of the differential equation $x\frac{dy}{dx} - y = 2x^2$ is

(A) $e^{-x}$

(B) $e^{-y}$

(C) $\frac{1}{x}$

(D) $x$

Answer:

The given equation is
$x\frac{dy}{dx} - y = 2x^2$
We can rewrite it as
$\frac{dy}{dx}-\frac{y}{x}= 2x$
Now,
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = \frac{-1}{x} \ and \ Q = 2x$
Now,
$I.F. = e^{\int pdx} = e^{\int \frac{-1}{x}dx}= e^{\int -\log x }= x^{-1}= \frac{1}{x}$
Therefore, the correct answer is (C)

Question 19: The Integrating Factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)$ is

(A) $\frac{1}{{y^2 -1}}$

(B) $\frac{1}{\sqrt{y^2 -1}}$

(C) $\frac{1}{{1 - y^2 }}$

(D) $\frac{1}{\sqrt{1 - y^2 }}$

Answer:

The given equation is
$(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)$
We can rewrite it as
$\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}$
It is $\frac{dx}{dy}+px= Q$ type of equation where $p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}$
Now,
$I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}$
Therefore, the correct answer is (D).

Question 1: Indicate Order and Degree.

(i) $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$

Answer:

The given function is
$\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$
We can rewrite it as
$y''+5x(y')^2-6y = \log x$
Now, it is clear from the above that the highest order derivative present in the differential equation is $y''$

Therefore, the order of the given differential equation $\frac{d^2y}{dx^2} + 5x \left (\frac{dy}{dx} \right )^2-6y = \log x$ is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y ', and the power raised to y '' is 1
Therefore, its degree is 1

Question 1: Indicate Order and Degree.

(ii) $\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$

Answer:

The given function is
$\left(\frac{dy}{dx} \right )^3 - 4\left(\frac{dy}{dx} \right )^2 + 7y = \sin x$
We can rewrite it as
$(y')^3-4(y')^2+7y=\sin x$
Now, it is clear from the above that the highest order derivative present in the differential equation is y'.
Therefore, the order of the given differential equation is 1
Now, the given differential equation is a polynomial equation in its derivatives, and the power raised to y ' is 3
Therefore, its degree is 3

Question 1: Indicate Order and Degree.

(iii) $\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$

Answer:

The given function is
$\frac{d^4 y}{dx^4} - \sin\left(\frac{d^3y}{dx^3} \right ) = 0$
We can rewrite it as
$y''''-\sin y''' = 0$
Now, it is clear from the above that the highest order derivative present in the differential equation is y''''

Therefore, the order of the given differential equation is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, its degree is not defined

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) $xy = ae^x + be^{-x} + x^2\qquad :\ x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy +x^2 -2 =0$

Answer:

Given,

$xy = ae^x + be^{-x} + x^2$

Now, differentiating both sides w.r.t. x,

$x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x$

Again, differentiating both sides w.r.t. x,

$\\ (x\frac{d^2y}{dx^2} + \frac{dy}{dx}) + \frac{dy}{dx} = ae^x + be^{-x} + 2 $
$\\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = ae^x + be^{-x} + 2 $
$\\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy -x^2 + 2$
$ \\ \implies x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 + 2$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(ii) $y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Answer:

Given,

$y = e^x(a\cos x + b \sin x )$

Now, differentiating both sides w.r.t. x,

$\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} = -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} $
$ \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iii) $y= x\sin 3x \qquad : \ \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Answer:

Given,

$y= x\sin 3x$

Now, differentiating both sides w.r.t. x,

$y= x\sin 3x \frac{dy}{dx} = x(3\cos 3x) + \sin 3x$

Again, differentiating both sides w.r.t. x,

$\\ \frac{d^2y}{dx^2} = 3x(-3\sin 3x) + 3\cos 3x + 3\cos 3x \\ = -9y + 6\cos 3x $
$ \implies \frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

Therefore, the given function is the solution of the corresponding differential equation.

Question 2: Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(iv) $x^2 = 2y^2\log y\qquad : \ (x^2 + y^2)\frac{dy}{dx} - xy = 0$

Answer:

Given,

$x^2 = 2y^2\log y$

Now, differentiating both sides w.r.t. x,

$\\ 2x = (2y^2.\frac{1}{y} + 2(2y)\log y)\frac{dy}{dx} = 2(y + 2y\log y)\frac{dy}{dx} $
$ \implies \frac{dy}{dx} = \frac{x}{y(1+ 2\log y)}$

Putting $\frac{dy}{dx}\ and \ x^2$ values in LHS

$\\ (2y^2\log y + y^2)\frac{dy}{dx} - xy = y^2(2\log y + 1)\frac{x}{y(1+ 2\log y)} -xy \\ = xy - xy = 0 = RHS$

Therefore, the given function is the solution of the corresponding differential equation.

Question 3: Prove that $x^2 - y^2 = c (x^2 + y^2 )^2$ is the general solution of differential equation $(x^3 - 3x y^2 ) dx = (y^3 - 3x^2 y) dy$ , where c is a parameter.

Answer:

Given,

$
\begin{aligned}
& \left(x^3-3 x y^2\right) d x=\left(y^3-3 x^2 y\right) d y \\
& \Longrightarrow \frac{d y}{d x}=\frac{\left(x^3-3 x y^2\right)}{\left(y^3-3 x^2 y\right)}
\end{aligned}
$
Now, let $\mathbf{y}=\mathrm{vx}$

$
\Longrightarrow \frac{d y}{d x}=\frac{d(v x)}{d x}=v+x \frac{d v}{d x}
$
Substituting the values of $y$ and $y^{\prime}$ in the equation,

$
\begin{aligned}
& v+x \frac{d v}{d x}=\frac{\left(x^3-3 x(v x)^2\right)}{\left((v x)^3-3 x^2(v x)\right)} \\
& \Longrightarrow v+x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v} \\
& \Longrightarrow x \frac{d v}{d x}=\frac{1-3 v^2}{v^3-3 v}-v=\frac{1-v^4}{v^3-3 v} \\
& \Longrightarrow\left(\frac{v^3-3 v}{1-v^4}\right) d v=\frac{d x}{x}
\end{aligned}
$
Integrating both sides, we get,

$
\int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\log x+\log C^{\prime}
$
Now, $\int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=\int \frac{v^3}{1-v^4} d v-3 \int \frac{v d v}{1-v^4}$

$
\Rightarrow \int\left(\frac{v^3-3 v}{1-3 v^4}\right) d v=I_1-3 I_2 \text {, where } I_1=\int \frac{v^3}{1-v^4} d v \text { and } I_2=\int \frac{v d v}{1-v^4}
$

Let $1-v^4=\mathrm{t}$

$
\begin{aligned}
& \frac{d}{d v}\left(1-v^4\right)=\frac{d t}{d v} \\
& \Longrightarrow-4 v^3=\frac{d t}{d v} \\
& \Longrightarrow v^3 d v=-\frac{d t}{4}
\end{aligned}
$
Now,

$
\mathrm{I}_1=\int-\frac{\mathrm{dt}}{4}=-\frac{1}{4} \log \mathrm{t}=-\frac{1}{4} \log \left(1-\mathrm{v}^4\right)
$

and

$
I_2=\int \frac{v d v}{1-v^4}=\int \frac{v d v}{1-\left(v^2\right)^2}
$
Let $v^2=p$

$
\begin{aligned}
& \Rightarrow \frac{d}{d v}\left(v^2\right)=\frac{d p}{d v} \\
& \Rightarrow 2 v=\frac{d p}{d v}
\end{aligned}
$

Question 4: Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$

Answer:

The given equation is
$\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$
We can rewrite it as
$\frac{dy}{dx } =- \sqrt{\frac{1-y^2}{1-x^2}}$
$ \frac{dy}{\sqrt{1-y^2}}= \frac{-dx}{\sqrt{1-x^2}}$
Now, integrate on both sides
$\sin^{-1}y + C =- \sin ^{-1}x + C'$
$ \sin^{-1}y+\sin^{-1}x= C$
Therefore, the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1-x^2}} = 0$ is $\sin^{-1}y+\sin^{-1}x= C$

Question 5: Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A (1 - x - y - 2xy)$ , where A is parameter.

Answer:
Given,

$
\begin{aligned}
& \frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0 \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\frac{\mathrm{y}^2+\mathrm{y}+1}{\mathrm{x}^2+\mathrm{x}+1}\right) \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}=\frac{\mathrm{dx}}{\mathrm{x}^2+\mathrm{x}+1} \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^2+\mathrm{y}+1}+\frac{\mathrm{dx}}{\mathrm{x}^2+\mathrm{x}+1}=0
\end{aligned}
$
Integrating both sides,

$
\begin{aligned}
& \int \frac{d y}{y^2+y+1}+\int \frac{d x}{x^2+x+1}=C \\
& \Rightarrow \int \frac{d y}{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\int \frac{d y}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=C \\
& \Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{\mathrm{y}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]=C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\mathrm{C} \Rightarrow \tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{2 y+1}{\sqrt{3}} \cdot \frac{2 x+1}{\sqrt{3}}}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{\frac{2 x+2 y+2}{\sqrt{3}}}{1-\left(\frac{4 x y+2 x+2 y+1}{3}\right)}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3}}{2} C \\
& \Rightarrow \frac{\sqrt{3}(x+y+1)}{(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3}}{2} c\right)
\end{aligned}
$
Let $\tan \left(\frac{\sqrt{3}}{2} c\right)=B$

$
x+y+1=\frac{2 B}{\sqrt{3}}(1-x-y-2 x y)
$

Let $A=\frac{2 B}{\sqrt{3}}$,

$
x+y+1=A(1-x-y-2 x y)
$

Hence proved.

Question 6: Find the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$

Answer:

The given equation is
$\sin x \cos y dx + \cos x \sin y dy = 0.$
We can rewrite it as
$\frac{dy}{dx}= -\tan x\cot y$
$ \frac{dy}{\cot y}= -\tan xdx$
$ \tan y dy =- \tan x dx$
Integrate both tides
$\log |\sec y|+C' = -\log|sec x|- C''$
$ \log|\sec y | +\log|\sec x| = C$
$ \sec y .\sec x = e^{C}$
Now, by using boundary conditions, we will find the value of C
It is given that the curve passing through the point $\left(0,\frac{\pi}{4} \right )$
So,
$\sec \frac{\pi}{4} .\sec 0 = e^{C}$
$ \sqrt2.1= e^C$
$ C = \log \sqrt2$
Now,
$\sec y.\sec x= e^{\log \sqrt 2}$
$ \frac{\sec x}{\cos y} = \sqrt 2$
$ \cos y = \frac{\sec x}{\sqrt 2}$
Therefore, the equation of the curve passing through the point $\left(0,\frac{\pi}{4} \right )$ whose differential equation is $\sin x \cos y dx + \cos x \sin y dy = 0.$ is $\cos y = \frac{\sec x}{\sqrt 2}$

Question 7: Find the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ , given that $y = 1$ when $x = 0$ .

Answer:

The given equation is
$(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$
We can rewrite it as
$\frac{dy}{dx}= -\frac{(1+y^2)e^x}{(1+e^{2x})}$
$ \frac{dy}{1+y^2}= \frac{-e^xdx}{1+e^{2x}}$
Now, integrate both sides
$\tan^{-1}y + C' =\int \frac{-e^{x}dx}{1+e^{2x}}$
$\int \frac{-e^{x}dx}{1+e^{2x}}\\$
Put
$e^x = t \\ e^xdx = dt$
$\int \frac{dt}{1+t^2}= \tan^{-1}t + C''$
Put $t = e^x$ again
$\int \frac{-e^{x}dx}{1+e^{2x}} = -\tan ^{-1}e^x+C''$
Put this in our equation
$\tan^{-1}y = -\tan ^{-1}e^x+C\\ \tan^{-1}y +\tan ^{-1}e^x=C$
Now, by using boundary conditions, we will find the value of C
It is given that
y = 1 when x = 0
$\\ \tan^{-1}1 +\tan ^{-1}e^0=C$
$ \frac{\pi}{4}+\frac{\pi}{4}= C\\ C = \frac{\pi}{2}$
Now, put the value of C

$\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$
Therefore, the particular solution of the differential equation $(1 + e^ {2x} ) dy + (1 + y^2 ) e^x dx = 0$ is $\tan^{-1}y +\tan ^{-1}e^x=\frac{\pi}{2}$

Question 8: Solve the differential equation $ye^\frac{x}{y}dx = \left(xe^\frac{x}{y} + y^2 \right )dy\ (y \neq 0)$

Answer:

Given,

$ye^\frac{x}{y}dx = (xe^\frac{x}{y} + y^2)dy$

$\\ ye^\frac{x}{y}\frac{dx}{dy} = xe^\frac{x}{y} + y^2 \\ \implies e^\frac{x}{y}[y\frac{dx}{dy} -x] = y^2 \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} = 1$

Let $\large e^\frac{x}{y} = t$

Differentiating it w.r.t. y, we get,

$\\ \frac{d}{dy}e^\frac{x}{y} = \frac{dt}{dy} \\ \implies e^\frac{x}{y}.\frac{d}{dy}(\frac{x}{y}) = \frac{dt}{dy} \\ \implies \frac{e^\frac{x}{y}[y\frac{dx}{dy} -x]}{y^2} =\frac{dt}{dy}$

Thus, from these two equations, we get,

$\\ \frac{dt}{dy} = 1 \\ \implies \int dt = \int dy \\ \implies t = y + C$

$\Rightarrow e^{\frac{x}{y}}=y+C$

Question 9: Find a particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ , given that $y = -1$ , when $x = 0$ . (Hint: put $x - y = t$ )

Answer:

The given equation is
$(x - y) (dx + dy) = dx - dy,$
Now, integrate both sides
Put
$(x-y ) = t\\ dx - dy = dt$
Now, the given equation becomes
$dx+dy= \frac{dt}{t}$
Now, integrate both sides
$x+ y + C '= \log t + C''$
Put $t = x- y$ again
$x+y = \log (x-y)+ C$
Now, by using boundary conditions, we will find the value of C
It is given that
y = -1 when x = 0
$0+(-1) = \log (0-(-1))+ C\\ C = -1$
Now, put the value of C

$x+y = \log |x-y|-1\\ \log|x-y|= x+y+1$
Therefore, the particular solution of the differential equation $(x - y) (dx + dy) = dx - dy,$ is $\log|x-y|= x+y+1$

Question 10: Solve the differential equation $\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1\; \ (x\neq 0)$ .

Answer:

Given,

$\left[\frac{e^{-2\sqrt x}}{\sqrt x} - \frac{y}{\sqrt x} \right ]\frac{dx}{dy} = 1$

$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \\ & \Rightarrow \frac{d y}{d x}+\frac{y}{\sqrt{x}}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}\end{aligned}$

This equation is in the form of $\frac{d y}{d x}+p y=Q$

$
\begin{aligned}
& p=\frac{1}{\sqrt{x}} \text { and } Q=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} \\
& \text { Now, I.F. }=e^{\int p d x}=e^{\int \frac{1}{\sqrt{x}} d x}=e^{2 \sqrt{x}}
\end{aligned}
$

We know that the solution of the given differential equation is:

$\begin{aligned} & y(I . F .)=\int(Q \cdot F .) d x+C \\ & \Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int\left(\frac{\mathrm{e}^{-2 \sqrt{\mathrm{x}}}}{\sqrt{\mathrm{x}}} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}\right) \mathrm{dx}+C \\ & \Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=\int \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}+\mathrm{C} \\ & \Rightarrow \mathrm{ye}^{2 \sqrt{\mathrm{x}}}=2 \sqrt{\mathrm{x}}+C\end{aligned}$

Question 11: Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$ , given that $y = 0 \ \textup{when}\ x = \frac{\pi}{2}$ .

Answer:

The given equation is
$\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$
This is $\frac{dy}{dx} + py = Q$ type where $p =\cot x$ and $Q = 4xcosec x$ $Q = 4x \ cosec x$
Now,
$I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x$
Now, the solution of a given differential equation is given by the relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C$
$y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C$
$ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C$
Now, by using boundary conditions, we will find the value of C
It is given that y = 0 when $x= \frac{\pi}{2}$
at $x= \frac{\pi}{2}$
$0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C$
$ C = - \frac{\pi^2}{2}$
Now, put the value of C
$y\sin x= 2x^2-\frac{\pi^2}{2}$
Therefore, the particular solution is $y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)$

Question 12: Find a particular solution of the differential equation $(x+1)\frac{dy}{dx} = 2e^{-y} -1$ , given that $y = 0$ when $x = 0$

Answer:

The given equation is
$(x+1)\frac{dy}{dx} = 2e^{-y} -1$
We can rewrite it as
$\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\$
Integrate both sides
$\int \frac{e^ydy}{2-e^y}= \log |x+1|\\$
$\int \frac{e^ydy}{2-e^y}$
Put
$2-e^y = t\\ -e^y dy = dt$
$\int \frac{-dt}{t}=- \log |t|$
put $t = 2- e^y$ again
$\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|$
Put this in our equation
$\log |2-e^y| + C'= \log|1+x| + C''\\ \log (2-e^y)^{-1}= \log (1+x)+\log C\\ \frac{1}{2-e^y}= C(1+x)$
Now, by using boundary conditions, we will find the value of C
It is given that y = 0 when x = 0
at x = 0
$\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}$
Now, put the value of C
$\frac{1}{2-e^y} = \frac{1}{2}(1+x)$
$ \frac{2}{1+x}= 2-e^y\\ \frac{2}{1+x}-2= -e^y\\ -\frac{2x-1}{1+x} = -e^y\\ y = \log \frac{2x-1}{1+x}$
Therefore, the particular solution is $y = \log \frac{2x-1}{1+x}, x\neq-1$

Question 13: The general solution of the differential equation $\frac{ydx - xdy}{y} = 0$ is

(A) $xy = C$

(B) $x = Cy^2$

(C) $y = Cx$

(D) $y = Cx^2$

Answer:

The given equation is
$\frac{ydx - xdy}{y} = 0$
We can rewrite it as
$dx = \frac{x}{y}dy\\ \frac{dy}{y}=\frac{dx}{x}$
Integrate both sides
We will get
$\log |y| = \log |x| + C\\ \log \frac{y}{x} = C \\ \frac{y}{x} = e^C\\ \frac{y}{x} = C\\ y = Cx$
Therefore, the answer is (C)

Question 14: The general solution of a differential equation of the type $\frac{dx}{dy} + P_1 x = Q_1$ is

(A) $ye^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(B) $ye^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

(C) $xe^{\int P_1 dy} = \int \left(Q_1 e^{\int P_1 dy} \right )dy +C$

(D) $xe^{\int P_1 dx} = \int \left(Q_1 e^{\int P_1 dx} \right )dx +C$

Answer:

The given equation is
$\frac{dx}{dy} + P_1 x = Q_1$
And we know that the general equation of this type of differential equation is

$xe^{\int p_1dy} = \int (Q_1e^{\int p_1dy})dy+ C$
Therefore, the correct answer is (C)

Question 15: The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is

(A) $xe^y + x^2 = C$

(B) $xe^y + y^2 = C$

(C) $ye^x + x^2 = C$

(D) $ye^y + x^2 = C$

Answer:

The given equation is
$e^x dy + (y e^x + 2x) dx = 0$
We can rewrite it as
$\frac{dy}{dx}+y=-2xe^{-x}$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = 1 \ and \ Q = -2xe^{-x}$
Now,
$I.F. = e^{\int p dx }= e^{\int 1dx}= e^x$
Now, the general solution is
$y(I.F.) = \int (Q\times I.F.)dx+C$
$y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\ ye^x= \int -2xdx + C\\ ye^x=- x^2 + C\\ ye^x+x^2 = C$
Therefore, (C) is the correct answer.

Differential Equations Class 12 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Differential Equations Class 12 Maths Chapter 9 are provided in the links below.

• Differential Equations Class 12 Exercise 9.1
• Differential Equations Class 12 Exercise 9.2
• Differential Equations Class 12 Exercise 9.3
• Differential Equations Class 12 Exercise 9.4
• Differential Equations Class 12 Exercise 9.5
• Differential Equations Class 12 Miscellaneous Exercise

Class 12 Maths NCERT Chapter 9: Extra Question

Question:
Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x$, $y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to:

Solution:
We have the differential equation,

$\begin{array}{l}\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x \\ \Rightarrow \frac{d y}{d x}+3 \sec ^2 x y=\sec ^2 x\end{array} $

Compare it with the linear differential equation,

$\frac{dy}{dx}+Py=Q$

We get, $Q=\sec^2 x$

Use this in the formula,

$I.F =e^{\int Pd x}$

$I.F =e^{\int 3 \sec ^2 x d x}=e^{3 \tan x}$

The solution to the differential equation is given by:

$y \cdot e^{3\tan x}=\int e^{3 \tan x} \cdot \sec ^2 x d x+c$

$y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c$
$\begin{aligned} & \text { Also } f(0)=\frac{1}{3}+e^3 \\ & \Rightarrow\left(\frac{1}{3}+e^3\right)=\frac{1}{3}+c \\ & \Rightarrow c=e^3 \\ & \therefore y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^3 \\ & \text { Put } x=\frac{\pi}{4} \\ & y e^3=\frac{e^3}{3}+e^3 \Rightarrow y=\frac{4}{3}\end{aligned}$

Hence, the correct answer is $\frac{4}{3}$.

Differential Equations Class 12 Chapter 9: Topics

Students will explore the following topics in NCERT Class 12 Maths Chapter 9 Differential Equations:

• Introduction
• Basic Concepts
• Order of a differential equation
• Degree of a differential equation
• General and Particular Solutions of a Differential Equation
• Methods of Solving First-Order, First-Degree Differential Equations
• Differential equations with variables separable
• Homogeneous differential equations
• Linear differential equations

Differential Equations Class 12 Solutions: Important Formulae

• A separable differential equation is one where the variables can be separated, i.e., it can be written in the form:
  $
  \frac{d y}{d x}=g(x) \cdot h(y)
  $
  To solve it, separate the variables and integrate:
  $
  \frac{1}{h(y)} d y=g(x) d x
  $
• A linear first-order differential equation has the form:
  $
  \frac{d y}{d x}+P(x) \cdot y=Q(x)
  $
  To solve this, use the integrating factor, which is:
  $
  \mu(x)=e^{\int P(x) d x}
  $
  Multiply the entire equation by the integrating factor and then integrate.
• An exact differential equation is one that satisfies:
  $
  M(x, y) d x+N(x, y) d y=0
  $
  where $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.
  To solve, find a potential function $\phi(x, y)$ such that:
  $
  \frac{\partial \phi}{\partial x}=M(x, y) \quad \text { and } \quad \frac{\partial \phi}{\partial y}=N(x, y)
  $
• For an equation of the form:
  $
  \frac{d y}{d x}+P(x) y=Q(x)
  $
  The integrating factor is given by:
  $
  \mu(x)=e^{\int P(x) d x}$

Chapter Summary of NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations

This chapter describes the concepts for differential equations and their solutions through various techniques. Concepts such as order and degree of differential equations, methods of variable separation, homogeneous differential equations, linear differential equations, and applications of differential equations are discussed. The NCERT Solutions in this chapter are presented in a very simple way, and concept clarity is achieved by solving step-by-step. Students practice 98 textbook questions given in 6 exercises. This gives a lot of conceptual clarity and makes students ready for application-based problems. Practice of this chapter enhances accuracy, analytical ability, and confidence in solving the problems in the examination. Also, students benefit later in this chapter when they encounter topics of Calculus.

Expert Review of NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations

Differential Equations is a crucial chapter of Calculus which demands absolute clarity of all concepts and continuous practice, according to the qualified Mathematics teachers of Careers360. Once the methods of solving each type of problem are learnt systematically, students will be able to handle application-based problems confidently. To make learning easy, the solutions to each technique have been made simple by detailed elaboration and logical presentation of steps. To ensure quick execution of calculations, it is advised to revise the formulas and practice each question in the NCERT book. Daily practice in this chapter favors well for CBSE Board examination as well as the JEE (Main and Advanced).

What Extra Should Students Study Beyond the NCERT for JEE?

For JEE aspirants, it is important to go beyond NCERT. Below are some extra topics that can help you build a deeper understanding and handle challenging problems with confidence.

NCERT Class 12 Solutions - Chapter Wise

Students can access all the Maths solutions from the NCERT book from the links below.

Also, read,

• NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations
• NCERT Notes Class 12 Maths Chapter 9 Differential Equations

NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:

• NCERT Exemplar Class 12th Maths Solutions
• NCERT Exemplar Class 12th Physics Solutions
• NCERT Exemplar Class 12th Chemistry Solutions
• NCERT Exemplar Class 12th Biology Solutions
• NCERT Exemplar Class 12 Maths Solutions Chapter 9

NCERT solutions for class 12 Subject-Wise

Here are the subject-wise links for the NCERT solutions of class 12:

• NCERT Solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions Class-Wise

Given below are the class-wise solutions of the NCERT :

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12

Also, check,

• NCERT