NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

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.

This Story also Contains

Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 9
What will the students learn in NCERT Exemplar Class 12 Maths Solutions Chapter 9?
NCERT Exemplar Class 12 Maths Solutions
Important Topics To Cover For Exams From NCERT Exemplar Class 12 Maths Solutions Chapter 9

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

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.

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.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

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

Must Read NCERT Solution subject wise

Also Check NCERT Books and NCERT Syllabus here

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.

Also Read

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Biology - Check CBSE 12 Biology NCERT Chapterwise Solution

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Class 12

NCERT Solutions for Class 12 Maths (Updated 2023-24)

NCERT Syllabus for Class 6 to 12 (2023-24) - All Subjects PDF Download

NCERT Syllabus for Class 12 Maths 2023 - Download PDF

NCERT Syllabus for Class 12 Chemistry 2023 - Download PDF

NCERT Syllabus for Class 12 Biology 2023

NCERT Syllabus for Class 12 Physics 2023 - Download PDF

NCERT Syllabus for Class 12 2023 (All Subjects) - Download PDF Here

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Books for class 12 Chemistry 2023 - Download Pdf Here

NCERT Books for Class 12 Hindi 2023 - Download chapter wise PDFs here

NCERT Book for Class 12 Maths 2023 - Download PDF

NCERT Books for Class 12 Biology 2023 - Download PDF

NCERT Book for Class 12 Physics 2023 - Download PDF

Haloalkanes and haloarenes class 12th notes- free ncert class 12 Chemistry Chapter 10 Notes- Download PDF

Integrals 12th Notes - Free NCERT Class 12 Maths Chapter 7 Notes - Download PDF

Application of Derivatives Class 12th Notes - Free NCERT Class 12 Maths Chapter 6 Notes - Download PDF

NCERT Class 12th Maths Notes Chapter-wise - Download Free PDF

NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

NCERT Class 12th Biology Notes Chapter-wise - Download Free PDF

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Articles

CBSE Class 12 Date Sheet 2024 - CBSE 12th Exam Date for Science, Commerce and Arts

Dec 07, 2023

NBSE HSSLC Routine 2024 (Out)- Nagaland Board Class 12th Exam Date Here

Dec 06, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 12 Class 12 - Linear Programming

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 12 - Three Dimensional Geometry

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 12 - Differential Equations

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals

Dec 04, 2023

CBSE Class 12 Date Sheet 2024 - CBSE 12th Exam Date for Science, Commerce and Arts

Dec 07, 2023

NBSE HSSLC Routine 2024 (Out)- Nagaland Board Class 12th Exam Date Here

Dec 06, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 12 - Continuity and Differentiability

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 12 - Differential Equations

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 12 - Probability

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

Dec 04, 2023

NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 12 - Application of Derivatives

Dec 04, 2023

3 min ⋆ Dec 06, 2023

Study Help

What Are Exponential And Logarithmic Functions, And How Do They Work?

5 min ⋆ Dec 05, 2023

Upcoming School Exams

All India Sainik Schools Entrance Examination

Application Date:06 November,2023 - 15 December,2023

Exam Pattern

Answer Key

Result

National Institute of Open Schooling 10th examination

Application Date:20 November,2023 - 19 December,2023

Application Process

Exam Pattern

Mock Test

National Institute of Open Schooling 12th Examination

Application Date:20 November,2023 - 19 December,2023

Application Process

Exam Pattern

Admit Card

West Bengal Board 12th Examination

Exam Date:30 November,2023 - 14 December,2023

Result

Exam Pattern

Admit Card

National Means Cum-Merit Scholarship

Application Date:03 December,2023 - 18 December,2023

Application Process

Exam Pattern

Mock Test

View All School Exams

Certifications By Top Providers

Artificial Intelligence Projects

Via Great Learning

Full-Stack Web Development

Via Masai School

Master Data Management for Beginners

Via TCS iON Digital Learning Hub

Digital Banking Business Model

Via State Bank of India

Introduction to Web Development with HTML CSS JavaScript

Via IBM

Google Cloud IAM and Networking for AWS Professionals

Via Google

1110 courses

783 courses

328 courses

136 courses

Harvard University, Cambridge

98 courses

IBM

86 courses

IT & Software

Teaching and Academics

Programming and Development

Health, Fitness and Medicine

Business and Management

Engineering and Architecture

General Management

Financial Management

History

Environmental Studies

Cloud Computing

Law

Explore Top Universities Across Globe

University of Essex, Colchester

Wivenhoe Park Colchester CO4 3SQ

University College London, London

Gower Street, London, WC1E 6BT

The University of Edinburgh, Edinburgh

Old College, South Bridge, Edinburgh, Post Code EH8 9YL

University of Nottingham, Nottingham

University Park, Nottingham NG7 2RD

Lancaster University, Lancaster

Bailrigg, Lancaster LA1 4YW

Bristol Baptist College, Bristol

The Promenade, Clifton Down, Bristol BS8 3NJ

How to Apply for PhD After Bachelors?

7 min ⋆ Nov 06, 2023 19:11 PM IST

USMLE 2023 Step 1 - Dates, Application Form, Exam Pattern, Score

10 min ⋆ Nov 06, 2023 19:11 PM IST

Top MBA Colleges in Dubai (with Tuition fees) –List of Best Colleges

5 min ⋆ Nov 06, 2023 13:11 PM IST

IELTS Result 2023: Dates, How to Check Score, Download Scorecard

11 min ⋆ Oct 25, 2023 05:10 AM IST

Permanent residency in Australia after studies - Procedure,How to improve chances to apply

6 min ⋆ Oct 23, 2023 06:10 AM IST

Letter of Recommendation (LOR) for PhD Students (with Sample) - Need, Parts, Qualities

9 min ⋆ Oct 23, 2023 06:10 AM IST

Related E-books & Sample Papers

UP Board 11th English Syllabus 2024

31+ Downloads

CBSE Class 12th Psychology Question Paper 2023

213+ Downloads

CBSE Class 12th Entrepreneurship Question Paper 2023

33+ Downloads

CBSE Class 12th Sociology Question Paper 2023

46+ Downloads

CBSE Class 12th Geography Question Paper 2023

39+ Downloads

CBSE Class 12th Accountancy Question Paper 2023

102+ Downloads

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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 will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Read Complete Answer

khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

Read Complete Answer

Parvati Solanki 02 June,2022

View All

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

Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available

Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available

Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available

Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian.

2 Jobs Available

Product Manager

3 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

Resource Links for Online MBA

3 Jobs Available

Team Lead

2 Jobs Available

Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available

Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available

Procurement Manager

The procurement Manager is also known as Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available

Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available