NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

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NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:11 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 9 provides an understanding of equations that relate to one or more functions and their derivatives. In this continually changing world, describing how things change with respect to several factors is very important. The representation of this information is termed as a differential equation. A differential equation is a great way to express a set of information, but it can be hard to solve or formulate. One of the essential languages of Science is that of differential equations.

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NCERT exemplar Class 12 Maths chapter 9 solutions provided on this page would help you gain academic success, ensure an efficient and easy way of clearing doubts, aid in preparation for 12 board exams, and shape your perspective about how the world works. Also, read NCERT Class 12 Maths Solutions

Question:1

Find the solution of $\frac{dy}{dx}=2^{y-x}$

Answer:

Given
$\frac{dy}{dx}=2^{y-x}$
To find: Solution of the given differential equation
Rewrite the equation as,
$\frac{dy}{2^{y}}=\frac{dx}{2^{x}}\\$
Integrating on both sides,
$\int \frac{dy}{2^{y}}=\int \frac{dx}{2^{x}}\\ \\ \int \frac{dx}{a^{x}}=-\frac{a^{-x}}{In a}$
Formula:
$- \frac{2^{-y}}{In 2}=-\frac{2^{-x}}{In 2}+c$
Here c is some arbitrary constant
$2^{-x}-2^{-y}=c\;In\;2\\ 2^{-x}-2^{-y}=d$
d is also some arbitrary constant = c In2

Question:2

Find the differential equation of all non-vertical lines in a plane.

Answer:

To find: Differential equation of all non vertical lines

The general form of equation of line is given by y=mx+c where, m is the slope of the line

The slope of the line cannot be $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ for the given condition because if it is so, the line will become perpendicular wihout any necessity.
So,
$m\neq \frac{\pi}{2},m\neq \frac{3\pi}{2}$
Differentiate the general form of equation of line
$\frac{dy}{dx}=m$
Formula:
$\frac{d(ax)}{dx}=a$
Differentiating it again it becomes,
$\frac{d^{2}y}{dx^{2}}=0$
Thus we get the diferential euqation of all non vertical lines.

Question:3

Given that $\frac{dy}{dx}=e^{-2y}$ and y = 0 when x = 5. Find the value of x when y = 3.

Answer:

Given:
$\frac{dy}{dx}=e^{-2y}$
(5,0) is a solution of this equation
To find: Solution of the given differential equation
Rewriting the equation.
$\frac{dy}{e^{-2y}}=dx$
Integrate on both the sides,
$\int \frac{dy}{e^{-2y}}=\int dx\\ \Rightarrow \int e^{2y}dy=\int dx$
Formula:
$\int e^{ax} dx =\frac{1}{a}e^{ax}\\ \frac{e^{2y}}{2}=x+c$
Given (5,0) is a solution so to get c, satisfying these values
$\frac{1}{2}=5+c\\ c=-\frac{9}{2}\\$
Hence the solution is
e^2y=2x + 9
when y=3,
e²⁽³⁾ =2x + 9
e⁶=2x + 9
e⁶+ 9=2x
$\Rightarrow x=\frac{e^{6}+9}{2}$

Question:4

Solve the differential equation $(x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}$

Answer:

Given:
$(x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}$
To find: Solution of the given differential equation
Rewriting the equations as,
$\frac{dy}{dx}+\frac{2xy}{\left (x^{2}-1 \right )}=\frac{1}{\left (x^{2}-1 \right )^{2}}$
It is a first order liner differential equation Compare it with,
$\frac{dy}{dx}+p(x)y=q(x)\\ p(x)=\frac{2x}{\left (x^{2}-1 \right )}\\ q(x)=\frac{1}{\left (x^{2}-1 \right )^{2}}\\$
Calculate Integrating Factor,
$IF=e^{\int p(x)dx}\\ \\ IF=e^{\int \frac{2x}{x^{2}-1}dx}\\ \\ \Rightarrow \int \frac{2x}{(x^{2}-1)}dx=\int \frac{(x+1)+(x-1)}{(x^{2}-1)}dx\\ \int \frac{dx}{x-1}+\int \frac{dx}{x+1}=ln\left ( x+1 \right )+ln\left ( x-1 \right )\\ Formula:\;\int \frac{dx}{x}=ln\;x\\ IF=e^{ln(x^{2}-1)} \\IF=x^{2}-1$
Hence, solution of the differential equation is given by,
$y.(IF)=\int q(x).(IF)dx\\ y(x^{2}-1)=\int \frac{1}{(x^{2}-1)^{2}}(x^{2}-1)dx\\ y(x^{2}-1)=\int \frac{1}{(x^{2}-1)}dx\\ Formula: \int \frac{1}{(x^{2}-1)}dx=\frac{1}{2}\log\left ( \frac{x-1}{x+1} \right )\\ \frac{1}{(x^{2}-1)}dx=\frac{1}{2}\log\left ( \frac{x-1}{x+1} \right )+c$

Question:5

Solve the differential equation $\frac{dy}{dx}+2xy=y$

Answer:

$\frac{dy}{dx}+2xy=y$
To find: Solution of the given differential equation
$\int \frac{1}{x}{dx}=ln x+c\\ \\ \int x^{n}dx=\frac{x^{n+1}}{n+1}+c$
Rewriting the given equation as,
$\frac{dy}{dx}=y\left ( 1-2x \right )\\ \\ \frac{dy}{y}=\left ( 1-2x \right )dx$
Integrate on both the sides,
$\int \frac{dy}{y}=\int \left ( 1-2x \right )dx\\ \ln y =\left ( x-x^{2} \right )+\log c\\ \ln y-\ln c=x-x^{2}\\ \ln \frac{y}{c}=x-x^{2}\\ \frac{y}{c}=e^{x-x^{2}}\\ y=ce^{x-x^{2}}\\$

Question:6

Find the general solution of $\frac{dy}{dx}+ay =e^{mx}$

Answer:

Given:
$\frac{dy}{dx}+ay=e^{mx}\\$
It is a first order differential equation. Comparing it with,
$\frac{dy}{dx}+p(x)y=q(x)\\$
P(x) =a
Q(x)=e^xm
Calculating Integrating Factor
$IF=e^{\int p(x)dx}\\ IF=e^{\int a \;dx}\\ IF=e^{ax}$
Hence the solution of the given differential equation is ,
$y\left (IF \right )=\int q(x).(IF)dx\\ y.(e^{ax})=\int e^{mx}e^{ax} dx\\ y.(e^{ax})=\int e^{\left (m+a \right )x} dx\\ Formula: \int e^{ax}dx=\frac{1}{a}e^{ax}\\ y.(e^{ax})=\frac{\left (e^{(m+a)x} \right )}{m+a}+c$

Question:7

Solve the differential equation $\frac{dy}{dx}+1=e^{x+y}$

Answer:

$\frac{dy}{dx}+1=e^{x+y}$
To find: Solution of the given differential equation
Assume x+y=t
Differentiate on both sides with respect to x
$1+\frac{dy}{dx}=\frac{dt}{dx}$
Substitute
$\frac{dy}{dx}+1=e^{x+y}$ in the above equation
$e^{x+y}=\frac{dt}{dx}\\ e^{t}=\frac{dt}{dx}\\$
Rewriting the equation,
$dx=e^{-t}dt\\$
Integrate on both the sides,
$\int dx=\int e^{-t}dt\\ formula: \int e^{x}dx=e^{x}\\ x=-e^{-t}+c$
$Substituting \; t=x+y\\ x=-e^{-(x+y)}+c$
Is the solution of the differential equation

Question:8

Solve: $ydx - xdy=x^{2}ydx$

Answer:

Given:
$y dx - x dy = x^{2}ydx$
To find: solution of the differential equation
Rewriting the given equation,
$\int \frac{1-x^{2}}{x}dx=\int \frac{dy}{y}\\ \int \frac{1}{x}-x dx =\int \frac{dy}{y}\\ Formula: \int \frac{dx}{x}=\ln x\\ \int x^{n}dx=\frac{x^{n+1}}{n+1}\\ \ln x-\frac{x^{2}}{2}+\ln c=\ln y\\ -\frac{x^{2}}{2}=\ln y-\ln x-\ln c\\ -\frac{x^{2}}{2}=\ln \frac{y}{cx}\\ y=cxe^{-\frac{x^{2}}{2}}$

Question:9

Solve the differential equation $\frac{dy}{dx}=1+x+y^{2}+xy^{2}$ when y = 0, x = 0.

Answer:

Given:
$\frac{dy}{dx}=\left ( 1+x \right )\left ( 1+y^{2} \right )$ and (0,0) is solution of the equation
To find: solution of the differential equation
Rewriting the given equation as,
$\frac{dy}{1+y^{2}}=\left ( 1+x \right )dx$
Integrating on both the sides
$\int \frac{dy}{1+y^{2}}=\int \left ( 1+x \right )dx\\ arctan\; y=x+\frac{x^{2}}{2}+c\\ Formula:\; \int \frac{dy}{1+y^{2}}=\tan^{-1}y \\ \int x^{n}dx=\frac{x^{n+1}}{n+1}\\$
Substitute(0,0) to find c’s value
0+0=c
c=0
Hence, the solution is
$\tan^{-1}y=x+\frac{x^{2}}{2}\\ y=\tan\left ( x+\frac{x^{2}}{2} \right )$

Question:10

Find the general solution of $(x + 2y^{3}) {\frac{dy}{dx}=y}$

Answer:

Given:
$\left ( x+2y^{3} \right )\frac{dy}{dx}=y$
To find: Solution of the differential equation
Rewriting the equation as
$\frac{dx}{dy}=\frac{\left ( x+2y^{3} \right )}{y}\\\frac{dx}{dy}=\frac{x}{y}+2y^{2}\\ \frac{dx}{dy}-\frac{x}{y}=2y^{2}\\$
It is a first order linear differential equation
Comparing it with
$\frac{dx}{dy}+p(y)x=q(y)\\ p(y)=-\frac{1}{y}\\ q(y)=2y^{2}$
Calculation the integrating factor,
$IF=e^{\int p(y)dy}\\ IF=e^{\int -\frac{1}{y}dy}\\ formula\; \frac{dt}{t}=\ln t\\ IF=e^{-\ln y}=\frac{1}{y}$ Therefore, the solution of the differential equation is
$x.\left ( IF \right )=\int q(y).(IF)dy\\ \frac{x}{y}=\int 2y \; dy\\ Formula: \int x^{n}dx=\frac{x^{n+1}}{n+1}\\ \frac{x}{y}=y^{2}+c\\ x=y^{3}+cy$

Question:11

If y(x) is a solution of $\frac{2 + \sin x}{1 +y}\frac{dy}{dx}=-\cos x$ and y(0) = 1, then find the value of $y=\frac{\pi}{2}$

Answer:

Given:

$\frac{2+\sin x}{1+y} \frac{d y}{d x}=-\cos x$

To find: Solution of the differential equation

Rewriting the given equation as,

$\frac{d y}{1+y}=\frac{-\cos x d x}{2+\sin x}$

Integrating on both sides,

$\\ \int \frac{d y}{1+y}=\int \frac{-\cos x d x}{2+\sin x} \\ \ln (1+y)=\int \frac{-\cos x d x}{2+\sin x}$

Let sinx=t and cos xdx= dt
$\begin{array}{l} \text { formula: } \frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x} \\ \ln (1+\mathrm{y})=\int \frac{-\mathrm{dt}}{2+\mathrm{t}} \\ \text { Formula: } \int \frac{\mathrm{dt}}{\mathrm{t}}=\ln \mathrm{t} \end{array}$

ln(1+y)=-ln(2+t)+logc

ln(1+y)+ln(2++sinx)=logc

(1+y)(2+sinx)=c

When x=0 and y=1

$\begin{array}{l} 1=\frac{c}{2+\sin 0}-1 \\ 1+1=\frac{c}{2} \end{array}$

c=4

$\begin{array}{l} \Rightarrow \mathrm{y}=\frac{4}{2+\sin \mathrm{x}}-1 \\ \text { If } \mathrm{x}=\pi / 2 \text { then, } \\ \mathrm{y}=\frac{4}{2+\sin \frac{\pi}{2}}-1 \end{array}$

$\begin{array}{l} y=\frac{4}{2+1}-1 \\\\ y=\frac{4}{3}-1 \\\\ y=\frac{1}{3} \end{array}$

Question:12

If y(t) is a solution of $(1+t)\frac{dy}{dt}-ty=1$ and y(0) = –1, then show that $y(1)=-\frac{1}{2}$

Answer:

$(1+t)\frac{dy}{dt}-ty=1$ and (0,-1) is a solution
To find: Solution for the differential equation
Rewriting the given equation as,
$\frac{dy}{dt}-\frac{ty}{1+t}=\frac{1}{1+t}$
It is a first order linear differential equation
Comparing it with,
$\frac{dy}{dt}-p(t)y=q(t)\\ p(t)=-\frac{t}{1+t}\\ q(t)=\frac{1}{1+t}$
Calculation Integrating Factor
$IF=e^{\int \frac{t}{1+t}dt}\\ \\ IF=e^{\int \frac{-t+1}{1+t}dt}\\ \\ IF=e^{\int -1+\frac{1}{1+t}dt}\\ \\ IF=e^{-t+\ln(1+t)}=e^{-t}(1+t)\\ \\$
Hence the solution for the differential equation is,
$y.(IF)=\int q(t).(IF)dt\\ y(e^{-t}(1+t))=\int e^{-t}(1+t)\frac{1}{1+t}dt\\ y(e^{-t}(1+t))=\int e^{-t}dt\\ y(e^{-t}(1+t))=-e^{-t}+c\\ formula: \int e^{-t}dt=-e^{-t}$
Substitution (0,-1) to find the value of c
$-1=-1+c\\ c=0\\ y(e^{-t}(1+t))=-e^{-t}$
The solution therefore y(1) is
$y(1)=-\frac{1}{1+1}=-\frac{1}{2}$

Question:13

Form the differential equation having $y = (\sin^{-1}x)^{2} + A \cos^{-1}x + B$ , where A and B are arbitrary constant, as its general solution.

Answer:

Given:
$y=(\arcsin x)^{2}+A \arccos x+B$
To find: Solution of the differential equation
Differentiating on both the sides,
$\frac{dy}{dx}=2 \arcsin x\frac{1}{\sqrt{1-x^{2}}}-A\frac{1}{\sqrt{1-x^{2}}}\\ \sqrt{1-x^{2}}\frac{dy}{dx}=2 \arcsin x-A\\ Formula: \frac{d}{dx}\left ( \arcsin x \right )=\frac{1}{\sqrt{1-x^{2}}}\\\frac{d}{dx}\left ( \arccos x \right )=-\frac{1}{\sqrt{1-x^{2}}}$
Differntiate again on both the sides
$\sqrt{1-x^{2}}\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\frac{x}{\sqrt{1-x^{2}}}=2\frac{1}{\sqrt{1-x^{2}}}\\ \left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=2\\ Formula: \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\\$
Hence the solution is
$\left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}-2=0\\$

Question:14

Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

Answer:

To find: Differential equations of all circles which pass though origin and centre lies on x axis
Assume a point (0,k) on y-axis
Radius of the circle is
$\sqrt{0^{2}+k^{2}}=k$
General form of the equation of circle is,
$(x-a)^{2}+(y-b)^{2}=r^{2}$
Here a, c is the center and r is the radius.
Substituting the values in the above equation,
$(x-0)^{2}+(y-b)^{2}=k^{2}\\ x^{2}+y^{2}-2yk=0.......(i)$
Differentiate the equation with respect to x
$2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0\\ Formula:\frac{d}{dx}(x^{n})=nx^{n-1}\\ k=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$
Substituting the value of k in (i)
$x^{2}+y^{2}-2y\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}=0\\ \left ( x^{2}+y^{2} \right )\frac{dy}{dx}-2yx+2y^{2}\frac{dy}{dx}=0\\ \left ( x^{2}-y^{2} \right )\frac{dy}{dx}-2yx=0$

Question:15

Find the equation of a curve passing through origin and satisfying the differential equation $(1+x^{2})\frac{dy}{dx}+2xy=4x^{2}$

Answer:

Given:
$\left ( 1+x^{2} \right )\frac{dy}{dx}+2xy=4x^{2}$ and (0,0) is a solution to the curve

To find: Equation of the curve satisfying the differential equation
Rewrite the given equation
$\frac{dy}{dx}+\frac{2xy}{\left ( 1+x^{2} \right )}=\frac{4x^{2}}{1+x^{2}}$
Comparing with
$\frac{dy}{dx}+p(x)y=q(x)\\ p(x)=\frac{2x}{1+x^{2}}\\ q(x)=\frac{4x^{2}}{1+x^{2}}\\$
Calculating Integrating Factor
$IF=e^{\int p(x)dx}\\ IF=e^{\int \frac{2x}{1+x^{2}}dx}$
Calculating
$\int \frac{2x}{1+x^{2}}dx$
Assume
$1+x^{2}=t\\ 2x\; dx=dt\\ \\ \int \frac{dt}{t}=\ln(t)\\ Formula:\int \frac{dx}{x}=\ln x\\ Substituting \;t=1+x^{2}\\ \int \frac{2x}{1+x^{2}}dx=\ln(1+x^{2})\\ IF=e^{\ln(1+x^{2})}=(1+x^2)$
Hence the solution is
$y.(IF)=\int q(x).(IF)dx\\ y(1+x^{2})=\int \frac{4x^{2}}{1+x^{2}}(1+x^{2})dx\\ y(1+x^{2})=\frac{4}{3}x^{3}+c\\ Formula: \int x^{n}dx=\frac{x^{n+1}}{n+1}$
Satisfying (0,0) in the equation of the curve to find the value of c
0+0=c
c=0
therefore equation of the curve is
$y(1+x^{2})=\frac{4}{3}x^{3}\\ y=\frac{4x^{2}}{3(1+x^{2})}$

Question:16

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

Answer:

Given:
$x^{2}\frac{dy}{dx}=x^{2}+y^{2}+xy$
To find: solution for the differential equation
Rewriting the given equation as
$\frac{dy}{dx}=1+\frac{y^{2}}{x^{2}}+\frac{y}{x}$
Clearly it is a homogenous equation
Assume y=vx
Differentiate on both sides
$\frac{dy}{dx}=V+x\frac{dv}{dx}$
Substituting dy/dx in the equation
$1+\frac{y^{2}}{x^{2}}+\frac{y}{x}=V+x\frac{dV}{dx}\\ \\1+V^{2}+V=V+x\frac{dV}{dx}\\ \\1+V^{2}=x\frac{dV}{dx}\\\\ \frac{dx}{x}=\frac{dV}{1+V^{2}}$
Integrating on both the sides
$\int \frac{dx}{x}=\int \frac{dV}{1+V^{2}}\\ \ln x=\arctan V+c\\\\ Formula: \int \frac{dx}{x}=\ln x+c\\\int \frac{dV}{1+V^{2}}=\tan^{-1}V+c\\ $Substitute $v=\frac{y}{x}\\ \ln x=\tan^{-1}\frac{y}{x}+c$
is the solution for the differential equation

Question:17

Find the general solution of the differential equation $(1 + y^{2}) + (x - e^{\tan^{-1}y})\frac{dy}{dx}=0$

Answer:

Given
$(1+y^{2})+(x-e^{\arctan y})\frac{dy}{dx}=0$
To find: Solution of the given differential equation
Rewrite the given equation as,
$(1+y^{2})\frac{dx}{dy}+x-e^{\arctan y}=0\\ \frac{dx}{dy}+\frac{x}{(1+y^{2})}=\frac{e^{\arctan y}}{(1+y^{2})}$
It is a first order differential equation
Comparing it with
$\frac{dx}{dy}+p(y)x=q(y)\\\\ p(y)=\frac{1}{(1+y^{2})}\\\\ q(y)=\frac{e^{\arctan y}}{(1+y^{2})}$
Calculating Integrating Factor
$IF=e^{\int p(y)dy}\\ IF=e^{\int \frac{1}{1+y^{2}}dy}\\ Formula: \int \frac{1}{(1+y^{2})}dy=\arctan y\\IF=e^{\arctan y}$
Hence the solution of the given differential equation is
$x.(IF)=\int q(y).(IF)dy\\ x.(e^{\arctan y})=\int \frac{e^{\arctan y}}{(1+y^{2})}(e^{\arctan y })dy\\$ Assume $ (e^{\arctan y })=t$
Differentiate on both the sides
$\frac{e^{\arctan y }}{(1+y^{2})}dy=dt\\ x.t = \int tdt \\x.t = \frac{t^2}{2}+c \\$Substituting$ \; t\\ x.(e^{\tan^{-1}y})=\frac{e^{2\tan^{-1}y}}{2}+c$

Question:18

Find the general solution of $y^{2}dx + (x^{2} - xy + y^{2}) dy = 0.$

Answer:

Given:
$y^{2}dx+(x^{2}-xy+y^{2})dy=0$
To find: Solution for the given differential equation
Rewrite the given equation
$y^{2}\frac{dx}{dy}=xy-x^{2}-y^{2}\\ \frac{dx}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}$
It is a homogenous differential equation
Assume x=vy
Differentiating on both the sides
$\frac{dx}{dy}=v+y\frac{dv}{dy}$
Substitute dy/dx in the given equation
$v+y\frac{dv}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}$
Substitute v=x/y
$v+y\frac{dv}{dy}=v-1-v^{2}\\ y\frac{dv}{dy}=-1-v^{2}\\\\ \frac{dv}{1+v^{2}}=-\frac{dy}{y}$
Integrating on both the sides
$\int \frac{dv}{1+v^{2}}=-\int \frac{dy}{y}\\ \tan^{-1}v=-\ln\;y+c\\ Formula: \int \frac{dx}{x}=\ln x +c\\\int \frac{dv}{1+v^{2}} =\tan^{-1}v+c\\ Substituting \; v=\frac{x}{y}\\ \tan^{-1}\frac{x}{y}=-\ln y+c$

Question:19

Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].

Answer:

Given:
$(x+y)(dx-dy)=dx+dy$
To find: Solution of the given differential equation
Rewriting the given equation
$\left ( x+y \right )\left ( 1-\frac{dy}{dx} \right )=\left ( 1+\frac{dy}{dx} \right )$
Assume x+y=z
Differentiate on both sides with respect to x
$1+\frac{dy}{dx}=\frac{dz}{dx}$
Substituting the values in the equation
$z\left ( 1-\frac{dz}{dx}+1 \right )=\frac{dz}{dx}\\ 2z-z\frac{dz}{dx}-\frac{dz}{dx}=0\\ 2z=\left ( z+1 \right )\frac{dz}{dx}\\ dx=\left ( \frac{1}{2}+\frac{1}{2z} \right )dz$
Integrate on both the sides
$\int dx=\int \left ( \frac{1}{2}+\frac{1}{2z} \right )dz\\ x=\frac{z}{2}+\frac{1}{2}\ln z+c\\ formula:\int \frac{dx}{x}=\ln x$
Substitute v=xy
$x=\frac{x+y}{2}+\frac{1}{2}\ln(x+y)+\ln c\\ x-y-\ln(x+y)-\ln c=0\\ \ln(x+y)+\ln c=x-y\\ \ln c(x+y)=x-y\\ c(x+y)=e^{x-y}\\ x+y=1/c(e^{x-y})\\ x+y=d(e^{x-y})\\ where d=\frac{1}{c}\\$

Question:20

Solve: $2(y+3)-xy\frac{dy}{dx}=0$ given that y (1) = –2

Answer:

Given:
$2\left ( y+3 \right )-xy\frac{dy}{dx}=0$
To Find: Solution of the differential equation
$2\frac{dx}{x}=\frac{ydy}{y+3}\\ 2\frac{dx}{x}=\frac{\left [ \left ( y+3 \right ) -3\right ]dy}{y+3}\\ 2\frac{dx}{x}=dy-\frac{3dy}{y+3}$
Integrating on both sides
$\int 2\frac{dx}{x}=\int dy-\int \frac{3dy}{y+3}\\ 2\ln x=y-3\ln(y+3)+c\\ Formula: \int \frac{dx}{x}=\ln x$
Substitute (-2,1) to find value of c
$\\0=-2+c \\c=2 \\2 \ln x =y-3 \ln(y+3)+2 \\ 2 \ln x+3\ln(y+3)=y+2 \\2 \ln x+3\ln(y+3)=y+2 \\ \ln x^{2}+ \ln(y+3)^{3}=y+2 \\x^{2}(y+3)^{3}=e^{y+2}$

Question:21

Solve the differential equation $dy = \cos x(2 - y\; cosec \;x) dx$ given that y = 2 when $x=\frac{\pi}{2}$

Answer:

Given:
$dy=\cos x(2-y\; cosec \;x)dx$
$\left ( \frac{\pi}{2},2 \right )$ is a solution of the given differential equation
Rewriting the given equation
$\frac{dy}{dx}=2\cos x-y \cot x\\ \frac{dy}{dx}+y \cot x=2\cos x\\$
It is a first order differential equation
$p(x)=\cot x\\ q(x)=2 \cos x$
Calculate integrating factor
$IF=e^{\int p(x)dx}\\ IF=e^{\int \cot x dx}\\ IF=e^{\ln \sin x}\\ Formula: \int \cot x =\ln \sin x\\ IF=\sin x$
Therefore, the solution of the differential equation is
$y.(IF)=\int q(x).(IF)dx\\ y \sin x= 2\int \cos x \sin x dx\\ y \sin x =\int \sin 2x \;dx\\ y \sin x = -\frac{1}{2}\cos 2x +c$
Substituting $\left ( \frac{\pi}{2},2 \right )$ to find the value of c
$2= \frac{1}{2}+c\\ c=\frac{3}{2}$
Hence the solution is
$y \sin x=-\frac{1}{2}\cos 2x+\frac{3}{2}$

Question:22

Form the differential equation by eliminating A and B in Ax² + By² = 1

Answer:

Given :
Ax²+By²=1
To find: Solution of the differential equation

Differentiate with respect to x
$2Ax+2By\frac{dy}{dx}=0\\ Ax+By\frac{dy}{dx}=0....(i)\\ Formula: \frac{d}{dx}(x^{n})=nx^{n-1}\\ \frac{dy}{dx}=-\frac{Ax}{By}.....(ii)$
Differentiate the curve (i) again to get,
$A+B\left ( \frac{dy}{dx}\frac{dy}{dx}+y\frac{d^{2}y}{dx^{2}} \right )=0\\ -\frac{A}{B}=\left ( \left ( \frac{dy}{dx} \right )^{2}+y\frac{d^{2}y}{dx^{2}} \right )$
Substituting this in eq(i)
$\frac{dy}{dx}=-\frac{x}{y}\left ( \left ( \frac{dy}{dx} \right )^{2}+y\frac{d^{2}y}{dx^{2}} \right )\\ y\frac{dy}{dx}=-x \left ( \frac{dy}{dx} \right )^{2}-xy\frac{d^{2}y}{dx^{2}}\\ y\frac{dy}{dx}+x \left ( \frac{dy}{dx} \right )^{2}+xy\frac{d^{2}y}{dx^{2}}=0$

Question:23

Solve the differential equation (1+ y²) tan^-1x dx + 2y (1 + x²) dy = 0

Answer:

Given:
$(1+y^{2})\tan^{-1}x dx +2y(1+x^{2})dy=0$
To find: Solution for differential equation that s given
Rewriting the given equation as.
$(1+y^{2})\tan^{-1}x =-2y(1+x^{2})\frac{dy}{dx}\\ \frac{\tan^{-1}x}{(1+x^{2})}dx=-\frac{2y}{(1+y^{2})}dy$
Integrate on the both sides
$\int \frac{\tan^{-1}x}{(1+x^{2})}dx=-\int \frac{2y}{(1+y^{2})}dy$
For LHS
Assume tan^-1 =t
$\frac{1}{1+x^{2}}dx=dt\\ Formula: \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^{2}}\\ \frac{d}{dx}(x^{n})=nx^{n-1}$
For RHS
Assume 1+y²=z
2ydy=dz
Substituting and integrating on both the sides
$\int t\; dt=-\int \frac{dz}{z}\\ \frac{t^{2}}{2}=-\ln z+c\\ Formula: \int \frac{dx}{x}=\ln x \\\int x^{n}dx=\frac{x^{n+1}}{n+1}$
Substitute for t and z
Solution for the differential equation is
$\frac{\left ( \tan^{-1}x \right )^{2}}{2}=-\ln\left ( 1+y^{2} \right )+c\\ \frac{\left ( \tan^{-1}x \right )^{2}}{2}+\ln\left ( 1+y^{2} \right )=c\\$

Question:24

Find the differential equation of system of concentric circles with centre (1, 2).

Answer:

To find: Differential equation of concentric circles whose center is (1,2)
Equation of the curve is given by
(x-a)²+(y-b)²=k²
Where (a,b) is the center and k, radius.
Subsitute the values now,
(x-1)²+(y-2)²=k²
Differentiate with respect to x
$2(x-1)+2(y-2)\frac{dy}{dx}=0\\ (x-1)+(y-2)\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{(x-1)}{(y-2)}$

Question:25

Solve: $y+\frac{dy}{dx}(xy)=x(\sin x +\log x)$

Answer:

$y+\frac{dy}{dx}(xy)=x(\sin x +\log x)$
Now dx/dy (xy) refers to the differentiation of xy with respect to x
Using product rule
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}$
When we put it back originally in the differential equation given,
$\\ \Rightarrow y+y+x \frac{d y}{d x}=x(\sin x+\log x) \\ \Rightarrow x \frac{d y}{d x}+2 y=x(\sin x+\log x)$
Divide by x
$\Rightarrow \frac{d y}{d x}+\left(\frac{2}{x}\right) y=\sin x+\log x$
Compare $\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{2}{\mathrm{x}}\right) \mathrm{y}=\sin \mathrm{x}+\log _{\mathrm{x}}$ with $ \quad \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$$
We get
$\mathrm{P}=\frac{2}{\mathrm{x}} \text { and } \mathrm{Q}=\sin \mathrm{x}+\log \mathrm{x}$
The above equation is a linear differential equation with P and Q as functions of x
The first to find the solution of a linear differential equation is to find the integrating factor. $\Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{Pdx}}$$
$\\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \int \frac{1}{\mathrm{x}} \mathrm{dx}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \log \mathrm{x}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2} \\ \Rightarrow \mathrm{IF}=\mathrm{x}^{2}$
The solution of the linear differential equation is
$y(\mathrm{IF})=\int Q(\mathrm{IF}) \mathrm{d} \mathrm{x}+\mathrm{c}$$
Substituting values for Q and IF
$\Rightarrow y x^{2}=\int(\sin x+\log x) x^{2} d x+c$$
$\Rightarrow y x^{2}=\int x^{2} \sin x d x+\int x^{2} \log x d x+c \ldots(a)$
Find the integrals individually,
Using uv for integration
$\\ \Rightarrow \int_{U V} d x=u \int v d x-\int\left(u^{\prime} j v\right) d x \\ \Rightarrow \int x^{2} \sin x d x=x^{2}(-\cos x)-\int 2 x(-\cos x) d x$
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 \int x \cos x d x$$
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2\left(x \sin x-\int \sin x d x\right)$
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2(x \sin x-(-\cos x))$$
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 x \sin x+2 \cos x \ldots \text { (i) }$
Now
Use product rule
$\\ \Rightarrow \int x^{2} \log x d x=\log x\left(\frac{x^{3}}{3}\right)-\int\left(\frac{1}{x}\right)\left(\frac{x^{3}}{3}\right) d x \\ \Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{1}{3} \int x^{2} d x$
$\Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{x^{3}}{9} \ldots(i i)$
Substitute (i) and (ii) in (a)
$\Rightarrow \mathrm{yx}^{2}=-\mathrm{x}^{2} \cos \mathrm{x}+2 \mathrm{x} \sin \mathrm{x}+2 \cos \mathrm{x}+\left(\frac{\mathrm{x}^{3}}{3}\right) \log \mathrm{x}-\frac{\mathrm{x}^{3}}{9}+\mathrm{c}$
Divide by $x^{2}$$
$\Rightarrow y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^{2}}+\left(\frac{x}{3}\right) \log x-\frac{x}{9}+\frac{c}{x^{2}}$

Question:26

Find the general solution of $(1 + \tan y) (dx - dy) + 2xdy = 0.$

Answer:

$(1+\tan y)(d x-d y)+2 x d y=0$
$\Rightarrow \mathrm{dx}-\mathrm{dy}+\tan \mathrm{y} \mathrm{d} \mathrm{x}-\tan \mathrm{y} \mathrm{dy}+2 \mathrm{xdy}=0$$
Divide throughout by dy
$\\\Rightarrow \frac{d x}{d y}-1+\tan y \frac{d x}{d y}-\tan y+2 x=0$ \\$\Rightarrow(1+\tan y) \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}-(1+\tan y)+2 \mathrm{x}=0$$
Divide by (1+tany)
$\\\Rightarrow \frac{d x}{d y}-1+\frac{2 x}{1+\tan y}=0$ \\$\Rightarrow \frac{d x}{d y}+\left(\frac{2}{1+t a n y}\right) x=1$$
$\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{2}{1+\operatorname{tany}}\right) \mathrm{x}=1 \quad \frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}$$
Compare
We get
$\\ P=\frac{2}{1+\tan y}$ \\$Q=1$$
This is the linear differential equation with P and Q as functions of x
$\\\Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{Pdy}}$ \\$\Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2}{1+\tan \mathrm{y}} \mathrm{dy}}$$
Put $tany =\frac{\sin y}{\cos y}$$
$\Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2 \cos y}{\sin y+\cos y}} \mathrm{dy}$
Adding and subtracting siny in the numerator
$\\ \Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{\cos y+\sin y+\cos y-\sin y}{\sin y+\cos y}} d y \\ \Rightarrow \mathrm{IF}=e^{\int\left(1+\frac{\cos y-\sin y}{\sin y+\cos y}\right) d y} \\ \Rightarrow \mathrm{IF}=e^{\int 1 \mathrm{~d} y+\int \frac{\cos y-\sin y}{\sin y+\cos y} d y}$
Consider the integral $\int \frac{\cos y-\sin y}{\sin y+\cos y} d y$$
Let $\sin y+\cos y=t$$
Differentiate with respect to y
We get
$\\\frac{\mathrm{dt}}{\mathrm{dy}}=\operatorname{cosy}-\sin y$ \\$\mathrm{dt}=(\cos \mathrm{y}-\mathrm{sin} \mathrm{y}) \mathrm{d} \mathrm{y}$$
$\begin{aligned} &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\int \frac{1}{t} d t\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log t\\ &\text { Resubstitue }\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log (\sin y+\cos y)\\ &\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}+\log (\sin y+\cos y)} \end{aligned}$
$\\\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}} \times \mathrm{e}^{\text {log }(\text { siny }+\cos y)}$ \\$\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}}(\mathrm{sin} \mathrm{y}+\cos \mathrm{y})$$
The solution of the linear differential equation will be $x(\mathrm{IF})=\int \mathrm{Q}(\mathrm{IF}) \mathrm{d} \mathrm{y}+\mathrm{c}$$
Substitute values for Q and IF
$\\\Rightarrow x e^{y}(\sin y+\cos y)=\int(1) e^{y}(\sin y+\cos y) d y+c$ \\$\Rightarrow x e^{y}(\sin y+\cos y)=\int\left(e^{y} \sin y+e^{y} \cos y\right) d y+c$$
Put $\mathrm{e}^{\mathrm{y}}$ sin y$=t$$ and differentiate with respect to y
We get
$\frac{d t}{d y}=e^{y} \sin y+e^{y} \cos y$
Which means
$\mathrm{dt}=\left(\mathrm{e}^{\mathrm{y}} \sin \mathrm{y}+\mathrm{e}^{\mathrm{y}} \cos \mathrm{y}\right) \mathrm{d} \mathrm{y}+\mathrm{c}$
Hence
$\\ \Rightarrow x e^{y}(\sin y+\cos y)=\int d t+c \\ \Rightarrow x e^{y}(\sin y+\cos y)=t+c$
Substitute t again
$x e^{y}(\sin y+\cos y)=e^{y} \sin y+c$$
$\Rightarrow \mathrm{x}=\frac{\sin \mathrm{y}}{\sin \mathrm{y}+\cos \mathrm{y}}+\frac{\mathrm{c}}{\mathrm{e}^{\mathrm{y}}(\sin \mathrm{y}+\mathrm{cosy})}$

Question:27

Solve: $\frac{dy}{dx}=\cos(x +y)+\sin(x+y)$ [Hint: Substitute x + y = z]
Answer:

$\frac{d y}{d x}=\cos (x+y)+\sin (x+y)$
Using the given hint substitute x+y=z
$\Rightarrow \frac{\mathrm{d}(\mathrm{z}-\mathrm{x})}{\mathrm{dx}}=\cos \mathrm{z}+\sin \mathrm{z}$
Differentiate z- x with respect to x
$\\ \Rightarrow \frac{d z}{d x}-1=\cos z+\sin z \\ \Rightarrow \frac{d z}{d x}=1+\cos z+\sin z \\ \Rightarrow \frac{d z}{1+\cos z+\sin z}=d x$
Integrate
$\Rightarrow \int \frac{\mathrm{d} z}{1+\cos z+\sin z}=\int \mathrm{d} \mathrm{x}$
As we know
$\cos 2 z=2 \cos ^{2} z-1$$
And $\sin 2 z=2 \sin z \cos z$$
$\\ \Rightarrow \int \frac{d z}{1+2 \cos ^{2} \frac{z}{2}-1+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x \\ \Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x \\ \Rightarrow \int \frac{d z}{2 \cos \frac{z}{2}\left(\cos \frac{z}{2}+\sin \frac{z}{2}\right)}=x \\ \Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}\left(1+\frac{\sin \frac{z}{2}}{\cos \frac{2}{2}}\right)}=x$
$\int \frac{\sec^2 \frac{z}{2}}{2(1+\tan \frac{z}{2})}dz$
$1+\tan \frac{z}{2}=t$$
Differentiate with respect to z
We get
$\frac{d t}{d z}=\frac{\sec ^{2} \frac{z}{2}}{2}$
hence $\frac{\sec ^{2} \frac{z}{2} \mathrm{dz}}{2}=\mathrm{dt}$$
$\Rightarrow \int \frac{d t}{t}=x$$
$logt +c=x$$
Again substitute t
$\Rightarrow \log \left(1+\tan \frac{z}{2}\right)+c=x$
Similarly substitute z
$\Rightarrow \log \left(1+\tan \frac{x+y}{2}\right)+c=x$

Question:28

Find the general solution of $\frac{dy}{dx}-3y= \sin 2x$

Answer:

$\\ \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \\ \text { Compare } \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \text { , and } \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
We get, P= -3 and Q= sin2x
The equation is a linear differential equation where P and Q are functions of x
For the solution of the linear differential equation, we need to find Integrating factor,
$\\ \Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{P} \mathrm{dx}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{[(-3) \mathrm{d} \mathrm{x}}$
$\Rightarrow \mathrm{IF}=\mathrm{e}^{-3 \mathrm{x}}$$
$y(\mathrm{IF})=\int \mathrm{Q}(\mathrm{IF}) \mathrm{d} \mathrm{x}+\mathrm{c}$
The solution of the linear differential equation is
Substitute values for Q and IF
$\Rightarrow y e^{-3 x}=\int e^{-3 x} \sin 2 x d x \ldots(1)$
Let $I=\int e^{-3 x} \sin 2 x d x$$
If $\mathrm{u}(\mathrm{x})$ and $\mathrm{v}(\mathrm{x})$$ are two functions, then by integration by parts.
$\int \mathrm{uv}=\mathrm{u} \int \mathrm{v}-\int \mathrm{u}^{\prime} \int \mathrm{v}$
$\mathrm{v}=\sin 2 \mathrm{x}$ and $\mathrm{u}=\mathrm{e}^{-2 x}$$
after applying the formula we get,
$\\I= \int \mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x} \mathrm{dx}=\mathrm{e}^{-3 \mathrm{x}} \int \sin 2 \mathrm{x}dx-\int\left(\mathrm{e}^{-3 \mathrm{x}}\right)^{\prime}dx \int \sin 2 \mathrm{x}dx \\ I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}+\int 3 \mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}dx+\mathrm{c}$
Again, applying the above stated rule in $\int 3 \mathrm{e}^{-3 \mathrm{x} \frac{\cos 2 \mathrm{x}}{2} }\text { we get }$
$I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{2}\left[\mathrm{e}^{-3 \mathrm{x}} \frac{\sin 2 \mathrm{x}}{2}+\int 3 \mathrm{e}^{-3 \mathrm{x}} \frac{\sin 2 \mathrm{x}}{2}\right]+\mathrm{c}$
$I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{4}\mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x}-\frac{9I}{4}+\mathrm{c}$
$\frac{13I}{4}=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{4}\mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x}+\mathrm{c}$
$I=\frac{-1}{13}\mathrm{e}^{-3 \mathrm{x}}( 2\cos 2 \mathrm{x}+3 \sin 2 \mathrm{x})+\mathrm{c}$
Put this value in (1) to get
$\\ \text { ye }^{- 3 x}=\int e^{-3 x} \sin 2 x d x \\ \text { ye }^{-3 x}=\frac{-e^{-3 x}(2 \cos 2 x+3 \sin 2 x)}{13}+c \\ \Rightarrow y=-\frac{1}{13}(3 \sin 2 x+2 \cos 2 x)+c e^{3 x}$

Question:29

Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is $\frac{x^{2}+y^{2}}{2xy}$

Answer:

Slope of the tangent is given by $\frac{x^{2}+y^{2}}{2 x y}$$
Slope of the tangent of the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{d} \mathrm{x}$$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{xy}}$$
Put y=VX
$\Rightarrow \frac{\mathrm{d}(\mathrm{vx})}{\mathrm{dx}}=\frac{\mathrm{x}^{2}+\mathrm{v}^{2} \mathrm{x}^{2}}{2 \mathrm{vx}^{2}}$$
Using product rule differentiate vx
$\\\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}=\frac{1+\mathrm{v}^{2}}{2 \mathrm{v}}$ \\$\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}}{2 \mathrm{v}}-\mathrm{v}$ \\$\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}-2 \mathrm{v}^{2}}{2 \mathrm{v}}$ \\$\Rightarrow \frac{d v}{d x}=\left(\frac{1}{x}\right) \frac{1-v^{2}}{2 v}$$
$\Rightarrow \frac{2 v}{1-v^{2}} d v=\left(\frac{1}{x}\right) d x$
Integrate
$\Rightarrow \int \frac{2 v d v}{1-v^{2}}=\int\left(\frac{1}{x}\right) d x$
Put $1-\mathrm{v}^{2}=\mathrm{t}$$
$2 \mathrm{vdv}=-\mathrm{dt}$$
$\\ \Rightarrow \int \frac{-d t}{t}=\log x+c \\ \Rightarrow-\log t=\log x+c$
Resubstitute 1
$\Rightarrow-\log \left(1-v^{2}\right)=\log x+c$
Resubstitute v
$\Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x+c \ldots(a)$
The curve is passing through (2,1)
Hence (2,1) will satisfy the equation (a)
Put x=1 and y=2 in (a)
$\\\Rightarrow-\log \left(1-\frac{1^{2}}{2^{2}}\right)=\log 2+c$ \\$\Rightarrow-\log \left(\frac{4-1}{4}\right)=\log 2+\mathrm{c}$ \\$\Rightarrow-\log \left(\frac{3}{4}\right)=\log 2+\mathrm{c}$ \\$\Rightarrow-\log \left(\frac{3}{4}\right)-\log 2=c$ \\$\Rightarrow-\left(\log \left(\frac{3}{4}\right)+\log 2\right)=c$$
Use loga+logb=logab
$\Rightarrow-\log \frac{3}{2}=c$$
Put c in equation (a)
$\\ \Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x-\log \frac{3}{2} \\ \Rightarrow-\log \left(\frac{x^{2}-y^{2}}{x^{2}}\right)=\log x-\log \frac{3}{2} \\ \Rightarrow \log \left(\frac{x^{2}-y^{2}}{x^{2}}\right)^{-1}=\log x-\log \frac{3}{2} \\ \Rightarrow \log \left(\frac{x^{2}}{x^{2}-y^{2}}\right)-\log x=-\log \frac{3}{2} \\ \Rightarrow \quad \frac{3 x}{2\left(x^{2}-y^{2}\right)}=e^{0} \\ \Rightarrow 3 x=2 x^{2}-2 y^{2}$
$\begin{aligned} &\Rightarrow 2 y^{2}=2 x^{2}-3 x\\ &\Rightarrow \mathrm{y}=\sqrt{\frac{2 \mathrm{x}^{2}-3 \mathrm{x}}{2}}\\ &\text { Hence the equation of the curve is }&y=\sqrt{\frac{2 x^{2}-3 x}{2}}\\ \end{aligned}$

Question:30

Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve any point (x, y) is $\frac{y-1}{x^{2}+x}$
Answer:

Given: Slope of the tangent is $\frac{y-1}{x^{2}+x}$$
Slope of tangent of a curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{dx}$$
$\\ \Rightarrow \frac{d y}{d x}=\frac{y-1}{x^{2}+x} \\ \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x}$
Integrate
$\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}-1}=\int \frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)} \ldots(\mathrm{a})$ $\frac{1}{x(x+1)}$$
Use partial fraction for
$\\\Rightarrow \frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$ \\$\Rightarrow \frac{1}{x(x+1)}=\frac{A(x+1)+B x}{x(x+1)}$$
Equate the numerator
$A(x+1)+B x=1$$
Put x=0
A=1
Put x=-1
B=-1
Hence $\Rightarrow \frac{1}{x(x+1)}=\frac{1}{x}+\frac{-1}{x+1}$$
Hence equation (a) becomes,
$\\ \Rightarrow \int \frac{d y}{y-1}=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) \mathrm{dx} \\ \Rightarrow \int \frac{d y}{y-1}=\int \frac{1}{x} d x-\int \frac{1}{x+1} d x$
$\log (y-1)=\log x-\log (x+1)+c \ldots(b)$$
Now it is given that the curve is passing through (1,0)
Hence (1,0) will satisfy the equation (b)
Put x=1 and y=0 in b
When we put y=0 in equation b the result is $\log (-1)$$ which is undefined
hence, we must simplify equation (b) further
$\log (y-1)-\log x=-\log (x+1)+c$
using loga-logb=loga/b
$\Rightarrow \log \left(\frac{y-1}{x}\right)(x+1)=c$
Constant c must be taken as log c to eliminate undefined elements in the
equation.(log cand not any other terms because taking logc completely
eliminates the log terms and we don't have to worry about undefined terms
in the equation)
$\Rightarrow \log \left(\frac{y-1}{x}\right)(x+1)=\log c$
Eliminate log
$\Rightarrow\left(\frac{y-1}{x}\right)(x+1)=c \ldots(c)$
Substitute x=1 and y=0
$\Rightarrow\left(\frac{0-1}{1}\right)(1+1)=c$
c=-2
put back c=-2 in (c)
$\begin{aligned} &\Rightarrow\left(\frac{y-1}{x}\right)(x+1)=-2\\ &\text { Hence the equation of the curve is }(y-1)(x+1)=-2 x \end{aligned}$

Question:31

Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

Answer:

Abscissa refers to the x coordinate and ordinate refers to the y coordinate.
Slope of the tangent is the square of the difference of the abscissa and the ordinate.
Difference of the abscissa and ordinate is (x-y) and its square is $(x-y) ^2$
Hence the Slope of the tangent is $(x-y) ^2$
$\frac{d y}{d x}=(x-y)^{2}$
$\begin{aligned} &\text { Put } \quad x-y=z\\ &\Rightarrow \quad 1-\frac{d y}{d x}=\frac{d z}{d x}\\ &\Rightarrow \quad 1-\frac{d z}{d x}=\frac{dy}{dx}=z^{2}\\ &\Rightarrow \quad 1-z^{2}=\frac{d z}{d x}\\ &\Rightarrow \quad d x=\frac{d z}{1-z^{2}}\\ &\Rightarrow \quad \int d x=\int \frac{d z}{1-z^{2}} \end{aligned}$
$\\x=\frac{1}{2} \log \left|\frac{1+z}{1-z}\right|+C \\ x=\frac{1}{2} \log \left| \frac{1+x-y}{1-x+y}\right|+C$
The curve passes through the (0,0)
$\\ 0=\frac{1}{2} \log 1+C \\ C=0$
$\\ \Rightarrow \mathrm{e}^{2 \mathrm{x}}=\frac{1+\mathrm{x}-\mathrm{y}}{1-\mathrm{x}+\mathrm{y}} \\ \Rightarrow \mathrm{e}^{2 x}(1-\mathrm{x}+\mathrm{y})=(1+\mathrm{x}-\mathrm{y})$

Question:32

Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x,y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Answer:

Points on the y axis and x axis are namely A(0,a), B(b,0). The midpoint of AB is P(x,y).
The x coordinate of the points is given by the addition of the x coordinates of A and B divided by 2.
$\begin{aligned} &\Rightarrow x=\frac{b+0}{2}\\ &b=2 x\\ &\text { Similarly, for y coordinate }\\ &\Rightarrow y=\frac{0+a}{2}\\ &a=2 y \end{aligned}$
Therefore, the coordinates of A and B are (0,2y) and (2x,0) respectively.
AB is the tangent to curve where P is the point of contact.
Slope of the line given with two points $\left(x_{1,} y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ on it is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
Here $\left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right)$ are $(0,2 y)(2 x, 0)$$ respectively.
Slope of the tangent AB is
$\frac{0-2 y}{2 x-0}$$
Hence the slope of the tangent is -y/x
Slope of the tangent curve is given by,
$\\\frac{\mathrm{dy}}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=-\frac{\mathrm{dx}}{\mathrm{x}}$$
Integrate
$\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}}=-\int \frac{\mathrm{dx}}{\mathrm{x}}$$
$\\logy =-\log x+c$ \\logy+ $\log x=c$$
using logat logb=logab.
$\log x y=c$$
as given curve is passing through(1,a)
Hence (1,1) will satisfy the equation of the curve(a)
Putting $x=1, y=2$ in $(a)$$
$\\\log 1=c$ \\$c=0$$
put c back in (a)
$\\\log x y=0$ \\$x y=e^{0}$$
$x y=1$
Hence the equation of the curve is $x y=1$$

Question:33

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

Answer:

$x \frac{d y}{d x}=y(\log y-\log x+1)$
Using loga-logb =loga/b
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)$
Put y=v x
$\Rightarrow \frac{d(v x)}{d x}=\frac{v x}{x}\left(\log \frac{v x}{x}+1\right)$
Differentiate yx with respect to x using product rule
$\Rightarrow \frac{d v}{d x} x+v=v(\log v+1)$
$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{x}+\mathrm{v}=\mathrm{v} \log \mathrm{v}+\mathrm{v}$$
$\\ \Rightarrow \frac{d v}{d x} x=v \log v \\ \Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$
Now Integrate
$\Rightarrow \int \frac{\mathrm{dv}}{\text { vlogv }}=\int \frac{\mathrm{dx}}{\mathrm{x}}$$
Substitute log v =t
Differentiate with respect to v.
$\frac{d v}{v}=d t$$
$\Rightarrow \int \frac{d t}{t}=\log x+c$$
logt= logx + logc
Resubstitute value of t
log(log v)=log x + logc.
Resubstitute v
$\\ \Rightarrow \log \left(\log \frac{y}{x}\right)=\log x+\log c \\ \Rightarrow \log \frac{y}{x}=c x$
Therefore the solution of the differential equation is
$\log \frac{y}{x}=c x$

Question:34

The degree of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^2+\left(\frac{d y}{d x}\right)^2=x \sin \frac{d y}{d x}$ is:
A. 1
B. 2
C. 3
D. Not defined

Answer:

Degree of differential equation is defined as the highest integer power of the highest order derivative in the equation.
Here’s the differential equation
$\left(\frac{d^{2} y}{d x^{2}}\right)^2+\left(\frac{d y}{d x}\right)^2=x \sin \frac{d y}{d x}$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Differential means
$\frac{d y}{d x} \text { or } \frac{d^{2} y}{d x^{2}} \text { or } \ldots \frac{d^{n} y}{d x^{n}}$
The given differential equation is not a polynomial because of the term sin dy/dx and therefore degree of such a differential equation is not defined.
Option D is correct.

Question:35

The degree of the differential equation $\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}$ is:
A. 4
B. 3/4
C. not defined
D. 2

Answer:

Generally, for a polynomial degree is the highest power.
$\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}=\frac{d^{2} y}{d x^{2}}$
Differential equation is Squaring both the sides,
$\Rightarrow\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$
Now for the degree to exit the differential equation must be a polynomial in
some differentials.
The given differential equation is polynomial in differential is
$\frac{\mathrm{dy}}{\mathrm{dx}}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$
Degree of differential equation is the highest integer power of the highest order
derivative in the equation.
Highest derivative is
$\frac{d^{2} y}{d x^{2}}$$
There is only one term of the highest order derivative in the equation which is
$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}$$ Whose power is 2 hence the degree is 2
Option D is correct.

Question:36

The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}=0$ respectively, are
A. 2 and 4
B. 2 and 2
C. 2 and 3
D. 3 and 3

Answer:

The differential equation is
$\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}=0$
Order is defined as the number which represents the highest derivative in a differential equation.
$\frac{d^{2} y}{d x^{2}}$
Is the highest derivative in the given equation is second order hence the degree of the equation is 2 .
Integer powers on the differentials,
$\\ \Rightarrow\left(\frac{d y}{d x}\right)^{\frac{1}{4}}=-\frac{d^{2} y}{d x^{2}}-x^{\frac{1}{5}} \\ \Rightarrow\left(\frac{d y}{d x}\right)^{\frac{1}{4}}=-\left(\frac{d^{2} y}{d x^{2}}+x^{\frac{1}{5}}\right)$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here differentials means
$\frac{\mathrm{dy}}{\mathrm{dx}}$ or $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ or $\ldots \frac{\mathrm{d}^{\mathrm{n}} \mathrm{y}}{\mathrm{dx}^{\mathrm{n}}}$$
The given differential equation is polynomial in differentials
Degree of differential equation is the highest integer power of the highest
order derivative in the equation.
Observe that
$\left(\frac{d^{2} y}{d x^{2}}+x^{\frac{1}{5}}\right)^{4}$$
Of differential equation (a) the maximum power $\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}$ will be 4$
Highest order is $\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}$$ and highest power is 4
Degree of the given differential equation is 4 .
Hence order is 2 and degree is 4
Option A is correct.

Question:37

if $\mathrm{y}=\mathrm{e}^{-\mathrm{x}}(\mathrm{A} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x})$ then y is a solution of

$\\A. \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}=0$ \\\\$\mathrm{B}$. $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$ \\\\C. $\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0$ \\\\D. $\frac{d^{2} y}{d x^{2}}+2 y=0$$

Answer:

If $\mathrm{y}=\mathrm{f}(\mathrm{x})$$ is a solution of differential equation, then differentiating it will give the same differential equation.
Differentiate the differential equation twice. Twice because all the options have order as 2 and also because there are two constants A and B
$y=e^{-x}(A \cos x+B \sin x)$$
Differentiating using product rule
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{e}^{-\mathrm{x}}(\mathrm{Acosx}+\mathrm{B} \sin \mathrm{x})+\mathrm{e}^{-\mathrm{x}}(-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x})$
But $e^{-x}(A \cos x+B \sin x)=y$$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}+\mathrm{e}^{-\mathrm{x}}(-\mathrm{Asinx}+\mathrm{Bcosx})$
Differentiating again with respect to x,
$\\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-e^{-x}(-A \sin x+B \cos x)+e^{-x}(-A \cos x-B \sin x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-e^{-x}(-A \sin x+B \cos x)-e^{-x}(A \cos x+B \sin x)$
But $e^{-x}(A \cos x+B \sin x)=y$$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}}(-\mathrm{Asinx}+\mathrm{B} \cos \mathrm{x})-\mathrm{y}$
Also,
$\frac{d y}{d x}=-y+e^{-x}(-A \sin x+B \cos x)$
Means,
$\\ e^{-x}(-A \sin x+B \cos x)=\frac{d y}{d x}+y \\ \quad \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-\left(\frac{d y}{d x}+y\right)-y$
$\\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}-\mathrm{y} \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-2 \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y} \\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0$

Question:38

The differential equation for $\mathrm{y}=\mathrm{A} \cos \alpha \mathrm{x}+\mathrm{B} \sin \alpha \mathrm{x},$ where $\mathrm{A}$ and $\mathrm{B}$$ are arbitrary constants is
$\\A. \frac{d^{2} y}{d x^{2}}-\alpha^{2} y=0$ \\\\$\mathrm{B}$ $\frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0$ \\\\C. $\frac{d^{2} y}{d x^{2}}+\alpha y=0$ \\\\D. $\frac{d^{2} y}{d x^{2}}-\alpha y=0$$

Answer:

Let us find the differential equation by differentiating y with respect to x twice
Twice because we have to eliminate two constants $\mathrm{A}$ and $\mathrm{B}$ \\$ .
$\mathrm{y}=\mathrm{A} \cos \alpha \mathrm{x}+\mathrm{B} \sin \alpha \mathrm{x}$
Differentiating,
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{A} \alpha \sin \alpha \mathrm{x}+\mathrm{B} \alpha \cos \alpha \mathrm{x}$
Differentiating again
$\\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-A \alpha^{2} \cos \alpha x-B \alpha^{2} \sin \alpha x \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2}(A \cos \alpha x+B \sin \alpha x)$
$\\ \text { But } y=A \cos a x+B \sin a x \\ \quad \Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2} y \\ \Rightarrow \frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0$
Option B is correct.

Question:39

Solution of differential equation xdy – ydx = 0 represents:
A. a rectangular hyperbola
B. parabola whose vertex is at origin
C. straight line passing through origin
D. a circle whose centre is at origin

Answer:

$\begin{aligned} &\text { Lets solve the differential equation }\\ &\begin{array}{l} x d y-y d x=0 \\ x d y=y d x \\ \Rightarrow \frac{d y}{y}=\frac{d x}{x} \\ \log y=\log x+c \\ \log x-\log y=c \\ \text { Using } \log a-\log b=\log a / b \\ \quad \Rightarrow \log \frac{y}{x}=c \\ \Rightarrow \frac{y}{x}=e^{c} \\ y=xe^{c} \end{array} \end{aligned}$
$\mathrm{e}^{\mathrm{c}}$$ is constant because e is a constant and c is the integration constant let it be denoted as k hence
$\\\mathrm{e}^{\mathrm{c}}=\mathrm{k}$ \\$y=k x$$
$\mathrm{y}=\mathrm{kx}$$ is the equation of straight line and (0,0) satisfies the equation.
Option C is correct.

Question:40

Integrating factor of the differential equation $\cos x \frac{d y}{d x}+y \sin x=1$ is:
A. cosx
B. tanx
C. sec x
D. sinx

Answer:

Differential equation is
$\\ \cos x \frac{d y}{d x}+y \sin x=1 \\ \Rightarrow \frac{d y}{d x}+\frac{y \sin x}{\cos x}=\frac{1}{\cos x} \\ \Rightarrow \frac{d y}{d x}+(\tan x) y=\sec x$
Compare
$\frac{d y}{d x}+(\tan x) y=\sec x$
With
$\frac{d y}{d x}+P y=Q^{\prime}$ we get, $P=\tan x$ and $Q=\sec x$$
The IF integrating factor is given by
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{\sin \mathrm{x}}{\cos \mathrm{x}} \mathrm{dx}}$
Substitute $\cos x=t$$ hence
$\\\frac{\mathrm{dt}}{\mathrm{dx}}=-\sin \mathrm{x}$ \\$\sin x d x=-d t$$
Resubstitute the value of t
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}}$$
$\\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}} \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log {\sec \mathrm{x}}} = \sec x$

hence IF is sec x
Option C is correct.

Question:41

Solution of the differential equation $\tan y\sec^2x dx + \tan x \sec^2 ydy = 0$ is:
A. tanx + tany = k
B. tanx – tan y = k
C. $\frac{\tan x}{\tan y}= k$
D. tanx . tany = k

Answer:

The given differential equation is
$\tan y\sec^2x dx + \tan x \sec^2 ydy = 0$
Divide it by tanx tany
$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\operatorname{tany}} d y=0$
Integrate
$\Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=0$
Put tanx=t hence,
$\sec ^{2} x d x=d t $$$
Put tany =z hence
$\frac{d z}{d y}=\sec ^{2} y$
That is $\sec ^{2} y d y=d t$$
$\Rightarrow \int \frac{d t}{t}+\int \frac{d z}{z}=0$$
$\log t+\log z+c=0$$
Resubstitue t and z
$\log (\tan x)+\log (\tan y)+c=0$$
$using \log a+\log \mathrm{b}=\log \mathrm{ab}$
$\log ($ tanxtany $)=-\mathrm{c}$$
$\tan x \tan y $=e^{-c}$$
$e^{-c}$$ is constant because e is a constant and c is the integration constant let it be denoted as $\mathrm{e}^{-c}=\mathrm{k}\\$
$\tan x \tan y $=\mathrm{k}$$
Option D is correct.

Question:42

Family $y = Ax + A^3$ of curves is represented by the differential equation of degree:
A. 1
B. 2
C. 3
D. 4

Answer:

$y=A x+A^{3}$$
let us find the differential equation representing it so we have to eliminate
the constant A
Differentiate with respect to x
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{A}$
Put back value of A in y
$\Rightarrow \mathrm{y}=\frac{\mathrm{dy}}{\mathrm{dx}} \mathrm{x}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{3}$$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here the differentials mean
$\frac{\mathrm{dy}}{\mathrm{dx}}$ or $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ or $\ldots \frac{\mathrm{d}^{\mathrm{n}} \mathrm{y}}{\mathrm{dx}^{\mathrm{n}}}$$
The given differential equation is polynomial in differentials
$\frac{\mathrm{dy}}{\mathrm{dx}}$$
Degree of differential equation is the highest integer power of the highest order derivative in the equation.
Highest derivative is
$\frac{\mathrm{dy}}{\mathrm{dx}}$$
And highest power to it is 3 . Hence degree is 3 .
Option C is correct.

Question:43

Integrating factor of $\frac{xdy}{dx}-y=x^4-3x$ is:
A. x
B. logx
C. $\frac{1}{x}$
D. –x

Answer:

Given differential equation
$\Rightarrow x \frac{d y}{d x}-y=x^{4}-3 x$
Divide though by x
$\\ \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{3}-3 \\ \Rightarrow \frac{d y}{d x}+\left(-\frac{1}{x}\right) y=x^{3}-3$
Compare
$\frac{\mathrm{d} y}{\mathrm{~d} x}+\left(-\frac{1}{x}\right) \mathrm{y}=\mathrm{x}^{3}-3 \\ \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
We get
$\mathrm{P}=-\frac{1}{\mathrm{x}} \text { and } \mathrm{Q}=\mathrm{x}^{3}-3$
The IF integrating factor is given by el $^{\mathrm{iPdx}}$
$\\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}} \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\log \mathrm{x}}$
$\begin{aligned} &\Rightarrow e^{\int P d x}=e^{\log x^{-1}}\\ &\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log _{\mathrm{x}}^{1}}\\ \\ &\text { Hence the IF integrating factor is } &\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\frac{1}{\mathrm{x}}\end{aligned}$
Option C is correct.

Question:44

Solution of $\frac{d y}{d x}-y=1, y(0)=1$ is given by
A. $xy = -e^x$

B. $xy = -e^{-x}$
C. $xy = -1$
D. $y = 2 e^x- 1$

Answer:

$\\ \frac{d y}{d x}-y=1$ \\$\Rightarrow \frac{d y}{d x}=1+y$ \\$\Rightarrow \frac{\mathrm{dy}}{1+\mathrm{y}}=\mathrm{dx}$$
Integrate
$\\\Rightarrow \int \frac{d y}{1+y}=\int d x\\$ $\log (1+x)=x+c$$
now it is given that y(0)=1 which means when x=0, y=1 hence substitute x=0 and y=0 in (a)
$\\\log (1+y)=x+c$ \\$\log (1+1)=0+c$ \\$\mathrm{c}=\log 2$$
put $\mathrm{c}=\log 2$$ back in (a)
$\\\log (1+y)=x+\log 2$ \\$\log (1+y)-\log 2=x$$
using $\log a-\log b=\log a / b$$
$\\\Rightarrow \log \frac{1+y}{2}=x$ \\$\Rightarrow \frac{1+y}{2}=e^{x}$ \\$1+y=2 e^{x}$ \\$y=2 e^{x}-1$$
Hence solution of differential equation is $\mathrm{y}=2 \mathrm{e}^{\mathrm{x}}-1$$
Option D is correct.

Question:45

The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$ when y(1) = 2 is:
A. none
B. one
C. two
D. infinite

Answer:

$\begin{aligned} &\begin{array}{l} \frac{d y}{d x}=\frac{y+1}{x-1} \\ \Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1} \end{array}\\ &\text { Integrate }\\ &\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}+1}=\int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}-1} \end{aligned}$
$\\\log (y+1)=\log (x-1)-\log c$ \\$\log (y+1)+\log c=\log (x-1)$$
using $\log a+\log \mathrm{b}=\log \mathrm{ab}$$
$\\\log _{0} c(y+1)=\log (x-1)$ \\$\Rightarrow \frac{x-1}{y+1}=c \ldots(a)$$
Now as given y(1)=2 which means when x=1, y=2 Substitute x=1 and y=2 in (a)
$\\\Rightarrow \frac{1-1}{2+1}=c$ \\$C=0$ \\$\Rightarrow \frac{x-1}{y+1}=0$ \\$x-1=0$$
So only one solution exists.
Option B is correct.

Question:46

Which of the following is a second order differential equation?
$\\A. \left(y^{\prime}\right)^{2}+x=y^{2}$ \\B. $y^{\prime} y^{\prime \prime}+y=\sin x$ \\C. $y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^{2}+y=0$ \\D. $y^{\prime}=y^{2}$$

Answer:

Order is defined as the number which defines the highest derivative in a differential equation
Second order means the order should be 2 which means the highest
derivative in the equation should be $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$$ or y Let's examine each of the option given
A. $\left(y^{\prime}\right)^{2}+x=y^{2}$$
The highest order derivative is $y^{\prime}$$ is in first order.
B. $y^{\prime} y^{\prime \prime}+y=\sin x$$
The highest order derivative is $y^{\prime \prime}$$ is in second order
C. $y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^{2}+y=0$$
The highest order derivative is $y^{\prime \prime \prime}$$ is in third order
D. $y^{\prime}=y^{2}$$
The highest order derivative is $y '$ is in first order
Option B is correct.

Question:47

Integrating factor of the differential equation $\left(1-x^{2}\right) \frac{d y}{d x}-x y=1$$ is:
A. -x
B. $\frac{x}{1+x^{2}}$$
C. $\sqrt{1-x^{2}}$$
D. $\frac{1}{2} \log \left(1-x^{2}\right)$$
Answer:

$\left(1-x^{2}\right) \frac{d y}{d x}-x y=1$$
Divide through by $\left(1-\mathrm{x}^{2}\right)$$
$\\ \Rightarrow \frac{d y}{d x}-\frac{x y}{1-x^{2}}=\frac{1}{1-x^{2}} \\ \Rightarrow \frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}}$
Compare
$\frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}} \quad\ and \ \ \frac{d y}{d x}+P y=Q$$
We get
$\mathrm{P}=\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}, \quad \mathrm{Q}=\frac{1}{1-\mathrm{x}^{2}}$
The IF factor is given by
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{-\mathrm{x}}{1-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}}$
Substitute $1-x^{2}=t$$ hence
$\frac{d t}{d x}=-2 x$$
Which means
$\\-x d x=\frac{d t}{2}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{\mathrm{dt}}{2 \mathrm{t}}}$ \\$\Rightarrow e^{\int \mathrm{Pdx}}=e^{\frac{1}{2} \int \frac{\mathrm{dt}}{t}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\frac{1}{2} \log t}$ \\$\Rightarrow e^{\int P d x}=e^{\log t^{\frac{1}{2}}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\mathrm{e}^{\log \sqrt{t}}$ \\$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\sqrt{\mathrm{t}}$$
Resubstitute
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\sqrt{1-\mathrm{x}^{2}}$$
Hence the IF integrating factor is $\sqrt{1-\mathrm{x}^{2}}$$
Option C is correct.

Question:48

$\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}$$ is the general solution of the differential equation:
A. $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$$
B. $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{x}^{2}}{1+\mathrm{y}^{2}}$$
C. $\left(1+x^{2}\right) d y+\left(1+y^{2}\right) d x=0$$
D. $\left(1+x^{2}\right) d x+\left(1+y^{2}\right) d y=0$$

Answer:

If $\mathrm{y}=\mathrm{f}(\mathrm{x})$$ is a solution of differential equation then differentiating it will give the same differential equation.
To find the differential equation differentiate with respect to x.
$$\tan ^{-1} x+\tan ^{-1} y=c$
$\\ \Rightarrow \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}\left(\frac{d y}{d x}\right)=0 \\\\ \left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0$
Option C is correct..

Question:49

The differential equation $\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{x}=\mathrm{c}$ represents:
A. Family of hyperbolas
B. Family of parabolas
C. Family of ellipses
D. Family of circles

Answer:

$\\ y \frac{d y}{d x}+x=c$ \\$\Rightarrow y \frac{d y}{d x}=c-x$ \\$y d y=(c-x) d x$$
integrate
$\\y d y=(c-x) d x$ \\$\Rightarrow \int y d y=\int(c-x) d x$ \\$\Rightarrow \int y d y=\int c d x-\int x d x$ \\$\Rightarrow \frac{y^{2}}{2}=c x-\frac{x^{2}}{2}+k$$
k is the integration constant
$\\\Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}=c x+k$ \\$\Rightarrow \frac{y^{2}+x^{2}}{2}=c x+k$$
This is the equation of circle because there is no ‘xy’ term and $x^2$ and $y^2$ have the same coefficient.
This equation represents the family of circles because for different values of c and k we will get different circles.
Option D is correct.

Question:50

The general solution of $e^x \cos y dx - e^x \sin y dy = 0$ is:
A. $e^x \cos y = k$
B. $e^x \sin y = k$
C. $e^x = k \cos y$
D. $e^x = k \sin y$

Answer:

$\\e^{x} \cos y d x-e^{x} \sin y d y=0$ \\$e^{x} \cos y d x=e^{x} \sin y d y$ \\$\Rightarrow d x=\frac{\sin y}{\cos y} d y$$
Integrate
$\Rightarrow \int \mathrm{dx}=\int \frac{\sin y}{\cos y} \mathrm{~d} y$$
substitute cosy =t hence
$\frac{\mathrm{dt}}{\mathrm{dy}}=-\sin \mathrm{y}$$
Which means sinydy=-dt
$\\\quad \Rightarrow x=\int \frac{-d t}{t}$ \\$x=-\log t+c$ \\$x+c=\log (\cos y)^{-1}$$
$\\ \Rightarrow x+c=\log \frac{1}{\cos y}$ \\$x+c=\log (\sec y)$ \\$e^{x+c}=\sec y$ \\$\mathrm{e}^{x } \mathrm{e}^{c}=\sec y$ \\$\Rightarrow \mathrm{e}^{\mathrm{x}}=\frac{1}{\mathrm{e}^{\mathrm{c}} \operatorname{cosy}}$ \\$e^{x} \cos y=e^{-c}$ \\$\mathrm{e}^{-\mathrm{c}}$ is constant because e is a constant and \\$\mathrm{c}$ is the integration constant let us it denote as $\mathrm{k}$ hence $\mathrm{e}^{-c}=\mathrm{k}$$
$e^{x} \cos y=k$
Option A is correct.

Question:51

The degree of the differential equation $\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{3}+6 y^{5}=0$$ is
(a) 1
(b) 2
(c) 3
(d) 5

Answer:

The answer is the option (a) 1 as the degree of a differential equation is the highest exponent of the order derivative.

Question:52

The solution of $\frac{d y}{d x}+y=e^{-x}, y(0)=0$$ is
(a) $y=e^{x}(x-1)$$
(b) $y=x e^{-x}$$
(c) $y=x e^{-x}+1$$
(d) $y=(x+1) e^{-x}$$

Answer:

The answer is the option (b) $y=x e^{-x}$$
Explanation: -
This is a linear differential equation.
On comparing it with $\frac{d y}{d x}+P y=Q$$ , we get
$P=1, Q=e^{-x}$
$\mathrm{IF}=e^{[P d x} e^{\int d x}=e^{x}$
So, the general solution is:
$\\y \cdot e^{x}=\int e^{-x} e^{x} d x+C$ \\$\Rightarrow \quad y \cdot e^{x}=\int d x+C$ \\$\Rightarrow \quad y \cdot e^{x}=x+C$$
Given that when x=0 and y=0
$\\\Rightarrow$ $0=0+C$ \\$\Rightarrow$ $C=0$$
Eq. (i) becomes $y \cdot e^{x}=x$$
$\Rightarrow \quad y=x e^{-x}$$

Question:53

Integrating factor of the differential equation $\mathrm{dy} / \mathrm{dx}+\mathrm{y} \tan \mathrm{x}-\sec \mathrm{x}=0$$
(a) $\operatorname{cos} x$$
(b) $\sec x$
(c) $e^{\cos x}$
(d) $e^{\sec x}$$

Answer:

The answer is the option (b) Sec x
Explanation: -
$\text { On comparing it with } \frac{d y}{d x}+P y=Q, \text { we get }$
$\\ P=\tan x, Q=\sec x \cdot \\\\ \text { I.F. }=e^{\int P d x}=e^{\int \operatorname{tan} x d x}=e^{(\log sec x)}=\sec x$

Question:54

The solution of the differential equation $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$$ is
(a) $y=\tan ^{-1} x$$
(b) $y-x=k(1+x y)$$
(c) $x=\tan ^{-1} y$$
(d) $\tan (x y)=k$$

Answer:

The answer is the option (b) y – x = k(1 + xy)
Explanation: -
$\begin{aligned} &\Rightarrow \quad \frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}\\ &\text { On integrating both sides, we get }\\ &\tan ^{-1} y=\tan ^{-1} x+C\\ &\Rightarrow \quad \tan ^{-1} y-\tan ^{-1} x=C\\ &\Rightarrow \quad \tan ^{-1}\left(\frac{y-x}{1+x y}\right)=C\\ &\Rightarrow \quad \frac{y-x}{1+x y}=\tan C\\ &\Rightarrow \quad y-x=\tan C(1+x y)\\ &\Rightarrow \quad y-x=k(1+x y), \text { where, } k=\tan C \end{aligned}$

Question:55

The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$$ is
(a) $\frac{x}{e^{x}}$$
(b) $\frac{e^{x}}{x}$$
(c) $x e^{x}$$
(d) $e^{x}$$

Answer:

The answer is the option (b) $\frac{e^{x}}{x}$$
Explanation: -
$\Rightarrow$ $\frac{d y}{d x}=\frac{1}{x}+\frac{y(1-x)}{x}$$
$\Rightarrow \quad \frac{d y}{d x}-\left(\frac{1-x}{x}\right) y=\frac{1}{x}$$
This is a linear differential equation.
On comparing it with $\frac{d y}{d x}+P y=Q,$$ we get
$\\ P=\frac{-(1-x)}{x}, Q=\frac{1}{x} \\ \mathrm{IF},=\int P d x=e^{-\int \frac{1-x}{x} d x} = \frac{e^x}{x}$

Question:58

The solution of $x \frac{d y}{d x}+y=e^{x}$$ is
(a) $y=\frac{e^{x}}{x}+\frac{k}{x}$$
(b) $y=x e^{x}+c x$$
(c) $y=x e^{x}+k$$
(d) $x=\frac{e^{y}}{y}+\frac{k}{y}$$

Answer:

The answer is the option (a) $y=\frac{e^{x}}{x}+\frac{k}{x}$$
Explanation: -
$\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=\frac{e^{x}}{x}$$
This is a linear differential equation. Dn comparing it with $\frac{d y}{d x}+P y=Q$, we get
$P=\frac{1}{x} \text { and } Q=\frac{e^{x}}{x}$
$\therefore \quad \mathrm{IF}_{0}=e^{\int \frac{1}{x} d x}=e^{(\log x)}=x$$
So, the general solution is:
$\\y \cdot x=\int \frac{e^{x}}{x} x d x$ \\$\Rightarrow \quad y \cdot x=\int e^{x} d x$ \\$\Rightarrow \quad \quad y \cdot x=e^{x}+k$ \\$\Rightarrow$ $y=\frac{e^{x}}{x}+\frac{k}{x}$$

Question:59

The differential equation of the family of curves $x^{2}+y^{2}-2 a y=0,$$ where a is arbitrary constant, is
(a) $\left(x^{2}-y^{2}\right) \frac{d y}{d x}=2 x y$$
(b) $2\left(x^{2}+y^{2}\right) \frac{d y}{d x}=x y$$
(c) $2\left(x^{2}-y^{2}\right) \frac{d y}{d x}=x y$$
(d) $\left(x^{2}+y^{2}\right) \frac{d y}{d x}=2 x y$$

Answer:

The answer is the option (a) $\left(x^{2}-y^{2}\right) \frac{d y}{d x}=2 x y$$
Given:
$x^{2}+y^{2}-2 a y=0.....(i)$
$\\2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0 \\\\ a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}$
$\\$Put the value of 'a' in Eq. (i) $ \\ x^{2}+y^{2}-2 y \frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}=0 \\\\ \left(x^{2}+y^{2}\right) \frac{d y}{d x}-2 x y-2 y^{2} \frac{d y}{d x}=0 \\ \\\left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0$$

Question:60

Family y = Ax + A³ of curves will correspond to a differential equation of order ,
(a) 3 (b) 2 (c) 1 (d) not defined.

Answer:

The answer is the option (c) 1.
Explanation: -
$\Rightarrow \quad \frac{d y}{d x}=A$$
Putting the value of A in Eq. (i), we gt
$y=x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{3}$
$\therefore \quad$ Order $=1$$

Question:61

The general solution of $\frac{d y}{d x}=2 x e^{x^{2}-y}$$ is
(a) $e^{x^{2}-y}=c$$
(b) $e^{-y}+e^{x^{2}}$$
(c) $e^{y}=e^{x^{2}}+c$$
(d) $e^{x^{2}+y}=c$$

Answer:

The answer is the option (c)
Explanation: -
$\begin{aligned} &\begin{array}{ll} \Rightarrow & e^{y} \frac{d y}{d x}=2 x e^{x^{2}} \\ \Rightarrow & \int e^{y} d y=2 \int x e^{x^{2}} d x \end{array}\\ &\text { Put } x^{2}=t \text { in } \mathrm{R} \text { . H.S. integral, we get }\\ &2 x d x=d t\\ &\begin{array}{ll} \Rightarrow & \int e^{y} d y=\int e^{t} d t \\ \Rightarrow & e^{y}=e^{t}+C \\ \Rightarrow & e^{y}=e^{x^{2}}+C \end{array} \end{aligned}$

Question:62

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is
(a) an ellipse (b) parabola (c) circle (d) rectangular hyperbola

Answer:

The answer is the option (d) Rectangular Hyperbola
Explanation: -
According to the question, $\frac{d y}{d x}=\frac{x}{y}$ $\Rightarrow \quad y d y=x d x$$
On integrating both sides, we get
$\frac{y^{2}}{2}=\frac{x^{2}}{2}+C$
$\Rightarrow \quad y^{2}-x^{2}=2 C,$$ which is an equation of rectangular hyperbola.

Question:63

The general solution of differential equation $\frac{d y}{d x}=e^{\frac{x^{2}}{2}}+x y$$ is
(a) $y=C e^{-x^{2} / 2}$$
(b) $y=C e^{x^{2} / 2}$$
(c) $y=(x+C) e^{x^{2} / 2}$$
(d) $y=(C-x) e^{x^{2} / 2}$$

Answer:

The answer is the option (c)
Explanation: -
$\Rightarrow \quad \frac{d y}{d x}-x y=e^{\frac{x^{2}}{2}}$$
This is a linear differential equation. On comparing it with $\frac{d y}{d x}+P y=Q,$$ we get
$P=-x, O=e^{x^{2} / 2}$
$\therefore \quad \mathrm{I.F.}=e^{\int -x d x}=e^{-x^{2} / 2}$$
So, the general solution is:
$\\\therefore \quad y \cdot e^{-x^{2} / 2}=\int e^{-x^{2} / 2} e^{x^{2} / 2} d x+C$ \\$\Rightarrow \quad y e^{-x^{2} / 2}=\int 1 d x+C$ \\$\Rightarrow \quad y e^{-x^{2} / 2}=x+C$ \\$\Rightarrow \quad y=(x+C) e^{x^{2} / 2}$$

Question:64

The solution of equation $(2 y-1) d x-(2 x+3) d y=0$$ is
(a) $\frac{2 x-1}{2 y+3}=k$$
(b) $\frac{2 y+1}{2 x-3}=k$$
(c) $\frac{2 x+3}{2 y-1}=k$$
(d) $\frac{2 x-1}{2 y-1}=k$$
Answer:

The answer is the option (c)
Explanation: -
$\begin{aligned} &\begin{array}{ll} \Rightarrow & (2 y-1) d x=(2 x+3) d y \\ \Rightarrow & \frac{d x}{2 x+3}=\frac{d y}{2 y-1} \end{array}\\ &\text { On integrating both sides, we get }\\ &\frac{1}{2} \log (2 x+3)=\frac{1}{2} \log (2 y-1)+\log C\\ &\begin{array}{ll} \Rightarrow & {[\log (2 x+3)-\log (2 y-1)]=2 \log C} \\ \Rightarrow & \log \left(\frac{2 x+3}{2 y-1}\right)=\log C^{2} \\ \Rightarrow & \frac{2 x+3}{2 y-1}=C^{2} \\ \Rightarrow & \frac{2 x+3}{2 y-1}=k, \text { where } K=C^{2} \end{array} \end{aligned}$

Question:65

The differential equation for which $y=a \cos x+b \sin x$$ is a solution, is
(a) $\frac{d^{2} y}{d x^{2}}+y=0$$
(b) $\frac{d^{2} y}{d x^{2}}-y=0$$
(c) $\frac{d^{2}}{d x^{2}}+(a+b) y=0$$
(d) $\frac{d^{2} y}{d x^{2}}+(a-b) y=0$$

Answer:

The answer is the option (a) $\frac{d^{2} y}{d x^{2}}+y=0$$
Explanation: -
On differentiating both sides w.r.t. x, we get $\frac{d y}{d x}=-a \sin x+b \cos x$$
Again, differentiating w.r.t. x, we get $\frac{d^{2} y}{d x^{2}}=-a \cos x-b \sin x$$
$\\\Rightarrow \quad \frac{d^{2} y}{d x^{2}}=-y$ \\$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+y=0$$

Question:66

The solution of $\frac{d y}{d x}+y=e^{-x}, y(0)=0$$ is
(a) $y=e^{-x}(x-1)$$
(b) $y=x e^{x}$$
(c) $y=x e^{-x}+1$$
(d) $y=x e^{-x}$$

Answer:

The answer is the option (d) $y=x e^{-x}$$
Explanation: -
$\frac{d y}{d x}+y=e^{-x}, y(0)=0$$
Here, $P=1$ and $Q=e^{-x}$ I.F. $=e^{\int \operatorname{ldx}}=e^{x}$$
$\\\therefore$ The general solution is $y \cdot e^{x}=\int e^{-x} \cdot e^{x} d x+C$ \\$\Rightarrow \quad y e^{x}=\int d x+C$ \\$\Rightarrow$ $y e^{x}=x+C$$
Given, when x=0 and y=0
$\\\Rightarrow$ $0=0+C$ \\$\Rightarrow \quad C=0$$
Eq. (i) reduces to $y \cdot e^{x}=x$ or $y=x e^{-x}$$

Question:67

The order and degree of the differential equation
$\left(\frac{d^{3} y}{d x^{3}}\right)^{2}-3 \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{4}=y^{4} \text { are }$
(a) 1,4
(b) 3,4
(c) 2,4
(d) 3,2

Answer:

Ans: - The answer is the option (d) 3, 2

Question:68

The order and degree of the differential equation $\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=\frac{d^{2} y}{d x^{2}}$ are
(a) $2, \frac{3}{2}$$
(b) 2,3
(c) 2,1
(d) 3,4

Answer:

Ans: -
The answer is the option (c) 2, 1.

Question:69

The differential equation of family of curves $y^{2}=4 a(x+a)$$ is
(a) $y^{2}=4 \frac{d y}{d x}\left(x+\frac{d y}{d x}\right)$$
(b) $2 y \frac{d y}{d x}=4 a$$
(c) $\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$$
(d) $2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$$

Answer:

Ans: - The answer is the option (d) $2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$$
Explanation: -
$y^{2}=4 a(x+a)$$
On differentiating both sides w.r.t. x, we get
$2 y \frac{d y}{d x}=4 a$
$\Rightarrow \quad \frac{1}{2} y \frac{d y}{d x}=a$$
On putting the value of a in Eq. (i), we get
$y^{2}=2 y \frac{d y}{d x}\left(x+\frac{1}{2} y \frac{d y}{d x}\right)$
$\Rightarrow \quad y^{2}=2 x y \frac{d y}{d x}+y^{2}\left(\frac{d y}{d x}\right)^{2}$$
$\Rightarrow \quad 2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$

Question:70

Which of the following is the general solution of $\frac{d^{2} y}{d x^{2}}-2\left(\frac{d y}{d x}\right)+y=0$ ?
(a) $y=(A x+B) e^{x}$$
(b) $y=(A x+B) e^{-x}$$
(c) $y=A e^{x}+B e^{-x}$$
(d) $y=A \cos x+B \sin x$$

Answer:

Ans: -
The answer is the option (a) $y=(A x+B) e^{x}$$
Explanation: -
$\begin{aligned} \quad \frac{d y}{d x} &=(A x+B) e^{x}+A e^{x}=(A x+A+B) e^{x} \\ \Rightarrow \quad & \frac{d^{2} y}{d x^{2}}=(A x+A+B) e^{x}+A e^{x}=(A x+2 A+B) e^{x} \\ \therefore \quad & \frac{d^{2} y}{d x^{2}}-2\left(\frac{d y}{d x}\right)+y =(A x+2 A+B) e^{x}-2(A x+A+B) e^{x}+(A x+B) e^{x} =0 \end{aligned}$

Question:71

General solution of $\frac{d y}{d x}+y \tan x=\sec x$ is
(a) $y \sec x=\tan x+C$$
(b) $y \tan x=\sec x+C$$
(c) $\tan x=y \tan x+C$$
(d) $x \sec x=\tan y+C$$

Answer:

Ans: - The answer is the option (a) y sec x = tan x + C
Explanation: -
Here, $P=\tan x, Q=\sec x$$
$\\\therefore$ I.F. $=e^{\int \tan x d x}=e^{\int \log \sec x}=\sec x$ \\$ \therefore$ The general solution is $y \sec x=\int{\sec } x \cdot \sec x+C$ \\$\Rightarrow$ $y \sec x=\int \sec ^{2} x d x+C$ \\$\Rightarrow$ $y \sec x=\tan x+C$$

Question:72

Solution of the differential equation $\frac{d y}{d x}+\frac{1}{x} y=\sin x$$ is
(a) $x(y+\cos x)=\sin x+C$$
(b) $x(y-\cos x)=\sin x+C$$
(c) $x y \cos x=\sin x+C$$
(d) $x(y+\cos x)=\cos x+C$$

Answer:

Ans: - The answer is the option (a) x(y + cos x) = sin x + C
Explanation: -
$\frac{d y}{d x}+\frac{1}{x} y=\sin x$$
Here, $P=\frac{1}{x}$ and $Q=\sin x$$
$\\\therefore \mathrm{I} . \mathrm{F},=e^{\int \frac{1}{x} d x}=e^{\log x}=x$ \\$\therefore$ The general solution is $y \cdot x=\int x \cdot \sin x d x+C$$
$\begin{aligned} &=-x \cos x-\int-\cos x d x \\ &=-x \cos x+\sin x+c \\ \Rightarrow x(y+\cos x) &=\sin x+C \end{aligned}$

Question:73

The general solution of differential equation $\left(e^{x}+1\right) y d y=(y+1) e^{x} d x$$ is
(a) $(y+1)=k\left(e^{x}+1\right)$$
(b) $y+1=e^{x}+1+k$$
(c) $y=\log \left\{k(y+1)\left(e^{x}+1\right)\right\}$$
(d) $y=\log \left\{\frac{e^{x}+1}{y+1}\right\}+k$$

Answer:

Ans: - The answer is the option (c) $y=\log \left\{k(y+1)\left(e^{x}+1\right)\right\}$$
Explanation: -
$\left(e^{x}+1\right) y d y=(y+1) e^{x} d x$$
$\\ \Rightarrow \frac{y d y}{y+1}=\frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow \int\left(1-\frac{1}{y+1}\right) d y=\int \frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow y-\log (y+1)=\log \left(e^{x}+1\right)+\log k \\ \Rightarrow y=\log (y+1)+\log \left(1+e^{x}\right)+\log k \\ \Rightarrow y=\log \left(k(1+y)\left(1+e^{x}\right)\right)$

Question:74

The solution of the differential equation $\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y}$$ is
(a) $y=e^{x-y}-x^{2} e^{-y}+c$$
(b) $e^{y}-e^{x}=\frac{x^{3}}{3}+c$$
(c) $e^{x}+e^{y}=\frac{x^{3}}{3}+c$$
(d) $e^{x}-e^{y}=\frac{x^{3}}{3}+c$$

Answer:

The answer is the option (b) $e^{y}-e^{x}=\frac{x^{3}}{3}+c$$
Explaination:
$\\ \frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \\\\ e^{y} d y=\left(e^{x}+x^{2}\right) d x \\\\ \int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x \\\\ e^{y}=e^{x}+\frac{x^{3}}{3}+C \\\\ e^{y}-e^{x}=\frac{x^{3}}{3}+C$

Question:75

The solution of the differential equation $\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}$ is$
(a) $y\left(1+x^{2}\right)=C+\tan ^{-1} x$$
(b) $\frac{y}{1+x^{2}}=C+\tan ^{-1} x$$
(c) $y \log \left(1+x^{2}\right)=C+\tan ^{-1} x$$
(d) $y\left(1+x^{2}\right)=C+\sin ^{-1} x$$

Answer:

Ans: - The answer is the option (a) $y\left(1+x^{2}\right)=C+\tan ^{-1} x$$
Explanation: -
$\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}$ is$
Here, $P=\frac{2 x}{1+x^{2}}$ and $Q=\frac{1}{\left(1+x^{2}\right)^{2}}$$
$\\ \therefore$ I.F. $=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$ \\$\therefore$ The general solution is$
$y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \frac{1}{\left(1+x^{2}\right)^{2}}+C$
$\\\Rightarrow \quad y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+C$ \\$\Rightarrow$ $y\left(1+x^{2}\right)=\tan ^{-1} x+C$$

Question:76

(i) The degree of the differential equation $\frac{d^{2} y}{d x^{2}}+e^{\frac{d y}{d x}}=0$ is
(ii) The degree of the differential equation $\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=x$$ is

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is

(iv) $\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$$ is an equation of the type

(v) General solution of the differential equation of the type $\frac{d y}{d x}+P y=Q$$ is given by
(vi) The solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}$$ is
(vii) The solution of $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$$ is
(viii) The solution of the differential equation $y d x+(x+x y) d y=0$$ is
(ix) General solution of $\frac{d y}{d x}+y=\sin x$$ is
(x) The solution of differential equation $\cot y d x=x d y$$ is
(xi) The integrating factor of $\frac{d y}{d x}+y=\frac{1+y}{x}$$ is

Answer:

(i) Given differential equation is
$\frac{d^{2} y}{d x^{2}}+e^{\frac{d y}{d x}}=0$
Degree of this equation is not defined as it cannot be expresses as polynomial of derivatives.
(ii) We have $\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=x$$
$\Rightarrow 1+\left(\frac{d y}{d x}\right)^{2}=x^{2}$$
So, degree of this equation is two.
(iii) Given that the general solution of a differential equation has three arbitrary constants. So we require three more equations to eliminate these three constants. We can get three more equations by differentiating given equation three times. So, the order of the differential equation is three.
(iv) We have $\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$$
The equation is of the type $\frac{d y}{d x}+P y=Q$$
Hence it is linear differential equation.
(v) We have $\frac{d x}{d y}+P_{1} x=Q_{1}$$
For solving such equation we multiply both sides by
So we get $e^{\int P_{1} d y}\left(\frac{d x}{d y}+P_{1} x\right)=Q_{1} e^{\int P_{1} d y}$$
$\\\Rightarrow \quad \frac{d x}{d y} e^{\int P_{1} d y}+P_{1} e^{\int P_{1}dy}=Q_{1} e^{\int P_{1}d y}$ \\\\$\Rightarrow \quad \frac{d}{d y}\left(x e^{\int P_{1} d y}\right)=Q_{1} e^{\int P_{1} dy}$ \\$\Rightarrow \quad \int \frac{d}{d y}\left(x e^{\int P_{1} d y}\right) d y=\int Q_{1} e^{P_{1} d y}dy$ \\$\Rightarrow \quad x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$$
This is the required solution of the given differential equation.
(vi) We have, $x \frac{d y}{d x}+2 y=x^{2}$$
$\frac{d y}{d x}+\frac{2 y}{x}=x$$
This equation of the form $\frac{d y}{d x}+P y=Q$.$
$\therefore \quad$ I.F. $=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^{2}$$
The general solution is
$y x^{2}=\int x \cdot x^{2} d x+C$
$\\\Rightarrow \quad y x^{2}=\frac{x^{4}}{4}+C$ \\$\Rightarrow \quad y=\frac{x^{2}}{4}+C x^{-2}$$

(vii) We have $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$$
$\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}$$
This equation is of the form $\frac{d y}{d x}+P y=Q$.$
$\therefore \quad \quad \quad \quad \mathrm{F} \cdot=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$
So, the general solution is:
$\begin{array}{ll} & y \cdot\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \frac{4 x^{2}}{\left(1+x^{2}\right)} d x+C \\ \Rightarrow & \left(1+x^{2}\right) y=\int 4 x^{2} d x+C \\ \Rightarrow & \left(1+x^{2}\right) y=\frac{4 x^{3}}{3}+C \\ \Rightarrow & y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}+C\left(1+x^{2}\right)^{-1} \end{array}$
(viii) We have, $y d x+(x+x y) d y=0$ $\Rightarrow \quad y d x+x(1+y) d y=0$ $
$\\ \Rightarrow$ $\frac{d x}{-x}=\left(\frac{1+y}{y}\right) d y$ \\$\int \frac{1}{x} d x=-\int\left(\frac{1}{y}+1\right) d y$ \\$\Rightarrow \quad \log x=-\log y-y+\log C$ \\$\Rightarrow \quad \log x+\log y-\log C=-y$ \\$\Rightarrow \quad \log \frac{x y}{C}=-y$ \\$\Rightarrow \quad \frac{x y}{C}=e^{-y}$$
$\Rightarrow \quad x y=C e^{-y}$$

(ix) We have, $\frac{d y}{d x}+y=\sin x$$
Which is of the form $\frac{d y}{d x}+P y=Q$$
$$$ \text { I.F. }=e^{\int 1 d x}=e^{x}$
So, the general solution is:
$\begin{array}{ll} & y \cdot e^{x}=\int e^{x} \sin x d x+C \\ \Rightarrow & y e^{x}=\frac{1}{2} e^{x}(\sin x-\cos x)+C \\ \Rightarrow & y=\frac{1}{2}(\sin x-\cos x)+C e^{-x} \end{array}$

(x) Given differential equation is $\cot y d x=x d y$$
$\Rightarrow $$ \begin{array}{l} \int \frac{1}{x} d x=\int \tan y d y \\ \log x=\log \sec y+\log C \end{array} $$$
$\\\Rightarrow$ $ \quad \log \frac{x}{\sec y}=\log C$ \\$\frac{x}{\sec y}=C$ \\$\Rightarrow$ $x=C \sec y$$

(xi) Given differential equation is
$\begin{array}{l} \frac{d y}{d x}+y=\frac{1+y}{x} \\ \frac{d y}{d x}+y=\frac{1}{x}+\frac{y}{x} \end{array}$
$\Rightarrow \quad \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x}$$
Which is linear differential equation.
$\therefore \quad \text { I.F. }=e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x-\log x}=e^{x} \cdot e^{-\log x}=\frac{e^{x}}{x}$

Question:77

i) Integrating factor of the differential of the form $\frac{d x}{d y}+P_{1} x=Q_{1}$$ is given by $e^{\int P_{1}d y}$$
(ii) Solution of the differential equation of the type $\frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}$$ is given by $x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$$
iii) Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x} f(x, y),$ where $f(x, y)$$ is a homogeneous function of zero degree is y=v x
(iv) Correct substitution for the solution of the differential equation of the type $\frac{d x}{d y} g(x, y)$ where g(x, y) is a homogeneous function of the degree zero is x=v y
(V) Number of arbitrary constants in the particular solution of a differential
equation of order two is two.
(vi)The differential equation representing the family of circles $x^{2}+(y-a)^{2}=$ $a^{2}$$ will be of order two.
(vii) The solution of $\frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3}$ is $y^{\frac{2}{3}}-x^{\frac{2}{3}}=c$$

(viii) Differential equation representing the family of curves $y=e^{x}(A \cos x+$$B \sin x$ ) is $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$$

ix) The solution of the differential equation $\frac{d y}{d x}=\frac{x+2 y}{x}$ is $x+y=k x^{2}$.$

x) Solution of $\frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right) \text { is } \sin \frac{y}{x}=C x$
xi) The differential equation of all non horizontal lines in a plane is $\begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}$

Answer:

i) Integrating factor of the differential of the form $\frac{d x}{d y}+P_{1} x=Q_{1}$$ is given by $e^{\int P_{1}d y}$$ . Hence given statement is true.

(ii) Solution of the differential equation of the type $\frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}$$ is given by $x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$$ .
Hence given statement is true.
iii) Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x} f(x, y),$ where $f(x, y)$$ is a homogeneous function of zero degree is y=v x.
Hence given statement is true.
(iv) Correct substitution for the solution of the differential equation of the type $\frac{d x}{d y} g(x, y)$ where g(x, y) is a homogeneous function of the degree zero is x=v y.
Hence given statement is true.
(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is Flase.
(vi) In thegiven equation $x^{2}+(y-a)^{2}=$ $a^{2}$$ the number of arbitrary constant is one. So the order order will be one.
Hence given statement is False.

(vii) $\frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3}$
$\frac{d y}{y\frac{1}{3}}=\left(\frac{dx}{x^\frac{1}{3}}\right)$
$\int \frac{d y}{y\frac{1}{3}}=\int \left(\frac{dx}{x^\frac{1}{3}}\right)$
$\begin{array}{l} \frac{3}{2} y^{2 / 3}=\frac{3}{2} x^{2 / 3}+C^{\prime} \\ y^{2 / 3}-x^{2 / 3}=C \end{array}$
Hence the given statement is true.
(viii) $y=e^{x}(A \cos x+$$B \sin x$ )$
$\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + e^x(A\sin x + B \cos x)$
$\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + y$
$\frac{d^2 y}{d x^2}=e^x(-A\sin x + B \cos x) + e^x(-A\cos x - B \sin x)+\frac{dy}{dx}$
$\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$ .
Hence the given statement is true.

ix) Given: $\frac{d y}{d x}=\frac{x+2 y}{x}$
$\frac{d y}{d x}-\frac{2}{x} y=1$
Compare with $\frac{d y}{d x}+P_{1} y=Q_{1}$$
Here $P_{1} = \frac{-2}{x}$ , $Q_{1} = 1$
$I.F. = e^{\int \frac{-2}{x}dx} = e^{\log \frac{1}{x^2}} = \frac{1}{x^2}$
General solution
$y. \frac{1}{x^2} = \int \frac{1}{x^2}dx$
$\frac{y}{x^2} = \frac{-1}{x}+c$
$y+x= cx^2$
Hence the given statement is true.
x) Given: $\frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right)$
$\frac{ d y}{d x}=\frac{y}{x}+ \tan \left(\frac{y}{x}\right)$
Let y =vx
$\frac{ d y}{d x}=v+x\frac{dv}{dx}$
$v+x\frac{dv}{dx} = v+\tan v$
$x\frac{dv}{dx} = \tan v$
$\int \frac{dv}{\tan v} = \int \frac{dx}{x}$
$\log \sin v = \log x + \log c$
$\sin v = xc$
$\sin \frac{y}{x} = cx$
Hence the given statement is true.
xi) Assume equation of a non-horizontal line in the plane
y = mx +c
$\frac{dy}{dx} = m$
$\frac{dx}{dy} = \frac{1}{m}$
$\begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}$
Hence the given statement is true.

Question:56

$y=ae^{mx}+be^{-mx}$ satisfies which of the following differential equation.
$\\a.\;\; \frac{dy}{dx}+my=0\\\\ b.\;\; \frac{dy}{dx}-my=0\\\\ c.\;\; \frac{d^{2}y}{dx^{2}}-m^{2}y=0\\\\ d.\;\; \frac{d^{2}y}{dx^{2}}+m^{2}y=0\\$

Answer:

given $y=ae^{mx}+be^{-mx}$
upon differentiation, we get $\frac{dy}{dx}=a.me^{mx}-b.me^{-mx}$
after differentiation again we get
$\frac{d^{2}y}{dx^{2}}=am^{2}e^{mx}-bm^{2}e^{-mx}\\\\ \Rightarrow \frac{d^{2}y}{dx^{2}}=m^{2}\left (ae^{mx}-be^{-mx} \right )\\\\ \Rightarrow \frac{d^{2}y}{dx^{2}}=m^{2}y\\\\ \Rightarrow \frac{d^{2}y}{dx^{2}}-m^{2}y=0\\$
Option c is correct

Question:57

The solution of the differential equation $\cos x \sin y \;dx+\sin x \cos y\; dy=0$ is
$\\(a)\frac{\sin x}{\sin y}=c\\ (b)\sin x \sin y = c\\ (c)\sin x +\sin y = c\\ (d)\cos x \cos y = c$

Answer:

$\\\cos x \sin y dx +\sin x \cos y dy=0\\ \Rightarrow \sin x \cos y dy=-\cos x \sin y dx\\ \Rightarrow \frac{\cos y}{sin y}dy=-\frac{\cos x}{sin x}dx\\ \Rightarrow \cot y dy =-\cot x dx$
Upon integration of both sides,
$\Rightarrow \int \cot y dy =-\int \cot x dx\\ \Rightarrow \log \left | \sin y \right |=-\log \left | \sin x \right |+\log c\\ \Rightarrow \log \left | \sin y \right |+\log \left | \sin x \right |=\log c\\ \Rightarrow \log \left | \sin y . \sin x \right |=\log c\\ \Rightarrow \sin y \sin x=c$

Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 9

Below is the list of topics which are covered in Class 12 Maths NCERT exemplar solutions chapter 9

Introductory Concepts
Ordinary differential equations
Order of a differential equation
Degree of a differential equation
General and particular solutions of a differential equation
Formation of a differential equation
Formation of a differential equation whose general solution is given
Formation of a differential equation that will represent a given family of curves
Methods of solving First order, First degree differential equations
Differential equations with separate variables
Homogeneous Differential equations
Linear differential equations

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What will the students learn in NCERT Exemplar Class 12 Maths Solutions Chapter 9?

These equations have a variety of uses in academic subjects like Physics, Chemistry, Biology, Geology, etc., which makes it important to acquire detailed learning of these equations.
A thorough understanding of differential equations is needed to solve Newton's laws of motion and cooling, the rate of spread of a pandemic or an epidemic, and also help measure market competition.
If understood well, NCERT exemplar solutions for Class 12 Maths chapter 9 would help you answer real-world questions like - How do you model an antibiotic-resistant bacteria's growth? How do you study ever-changing online purchasing trends? At what rate, a radioactive material decays? What is the trajectory of a biological cell motion? How the suspension system of a car works to give you a smooth ride? These equations are a way to describe many things in the universe and model nearly anything around us. Scientists and geniuses understand the world through differential equations; you can too.

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NCERT Exemplar Class 12 Maths Solutions

Chapter 1 Relations and Functions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 5 Continuity and Differentiability

Chapter 6 Applications of Derivatives

Chapter 7 Integrals

Chapter 8 Applications of Integrals

Chapter 10 Vector Algebra

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Maths Solutions Chapter 9

NCERT exemplar Class 12 Maths chapter 9 solutions revolve around using the mathematical tool of Differentiation, analyses properties like the intervals of increment and decrement, study the local maximum and minimum of functions that are quadratic.
In Class 12 Maths NCERT exemplar solutions chapter 9, you will be introduced to the concepts related to differential equations, types of these equations, general and specific solutions to solve them, the formation and production of these equations, different forms of the equations, and a wide range of applications of modelling real-life situations by applying these equations
In NCERT exemplar Class 12 Maths solutions chapter 9 pdf download, we would also look at the graphical aspects of differential equations, including a family of straight lines and curves, and have a look at the devised solutions and mathematical tools to solve the most complex equations over time.

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NCERT Exemplar Class 12 Solutions

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Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

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Frequently Asked Question (FAQs)

1. Are these solutions helpful in competitive exams?

Yes, these NCERT exemplar Class 12 Maths solutions chapter 9 can be highly useful in understanding the way the questions should be solved in entrance exams.

2. How to make use of these NCERT exemplar Class 12 Maths chapter 9 solutions?

These solutions can be used for both getting used to the chapter and its topics and to also get an idea about how to solve questions in exams.

3. What are the basic take away from the Class 12 Maths NCERT exemplar solutions chapter 9?

One can understand how to stepwise solve these questions through NCERT exemplar Class 12 Maths solutions chapter 9 and how the CBSE expects a student to solve in their final paper.

4. Who prepare these solutions to the maths chapter?

We have the best maths teachers onboard to solve the questions as per the students understanding and also CBSE standards. These teachers prepare the NCERT exemplar solutions for Class 12 Maths chapter 9.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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