NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

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# NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

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

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

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

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.

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.

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
e2y=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 $(x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}$

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

$\frac{dy}{dx}+2xy=y$
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 $\frac{dy}{dx}+ay =e^{mx}$

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)=exm
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 $\frac{dy}{dx}+1=e^{x+y}$

$\frac{dy}{dx}+1=e^{x+y}$
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: $ydx - xdy=x^{2}ydx$

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.

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 câ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 $(x + 2y^{3}) {\frac{dy}{dx}=y}$

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 $y=\frac{\pi}{2}$

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) = â1, then show that $y(1)=-\frac{1}{2}$

$(1+t)\frac{dy}{dt}-ty=1$ 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.

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.

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
$\sqrt{0^{2}+k^{2}}=k$
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 $(1+x^{2})\frac{dy}{dx}+2xy=4x^{2}$

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

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

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 $y^{2}dx + (x^{2} - xy + y^{2}) dy = 0.$

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 â dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].

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) = â2

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 $x=\frac{\pi}{2}$

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

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

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).

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

$y+\frac{dy}{dx}(xy)=x(\sin x +\log x)$
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 $(1 + \tan y) (dx - dy) + 2xdy = 0.$

$(1+\tan y)(d x-d y)+2 x d y=0$
$\Rightarrow \mathrm{dx}-\mathrm{dy}+\tan \mathrm{y} \mathrm{d} \mathrm{x}-\tan \mathrm{y} \mathrm{dy}+2 \mathrm{xdy}=0$
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

$\frac{d y}{d x}=\cos (x+y)+\sin (x+y)$
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 $\frac{dy}{dx}-3y= \sin 2x$

$\\ \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \\ \text { Compare } \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \text { , and } \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
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 $\frac{x^{2}+y^{2}}{2xy}$

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

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.

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.

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

$x \frac{d y}{d x}=y(\log y-\log x+1)$
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

Degree of differential equation is defined as the highest integer power of the highest order derivative in the equation.
Hereâ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

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 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

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

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

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 â 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

\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

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 â tan y = k
C. $\frac{\tan x}{\tan y}= k$
D. tanx . tany = k

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

$y=A x+A^{3}$
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. âx

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. $y = 2 e^x- 1$

$\\ \frac{d y}{d x}-y=1 \\\Rightarrow \frac{d y}{d x}=1+y \\\Rightarrow \frac{\mathrm{dy}}{1+\mathrm{y}}=\mathrm{dx}$
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

\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?
$\\A. \left(y^{\prime}\right)^{2}+x=y^{2} \\B. y^{\prime} y^{\prime \prime}+y=\sin x \\C. y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^{2}+y=0 \\D. y^{\prime}=y^{2}$

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. $\frac{1}{2} \log \left(1-x^{2}\right)$

$\left(1-x^{2}\right) \frac{d y}{d x}-x y=1$
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. $\left(1+x^{2}\right) d x+\left(1+y^{2}\right) d y=0$

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

$\\ y \frac{d y}{d x}+x=c \\\Rightarrow y \frac{d y}{d x}=c-x \\y d y=(c-x) d x$
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 âxyâ 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. $e^x = k \sin y$

$\\e^{x} \cos y d x-e^{x} \sin y d y=0 \\e^{x} \cos y d x=e^{x} \sin y d y \\\Rightarrow d x=\frac{\sin y}{\cos y} d y$
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 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) $y=(x+1) e^{-x}$

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) $e^{\sec x}$

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) $\tan (x y)=k$

The answer is the option (b) y â 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) $e^{x}$

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) $x=\frac{e^{y}}{y}+\frac{k}{y}$

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

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.

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) $e^{x^{2}+y}=c$

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

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) $y=(C-x) e^{x^{2} / 2}$

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) $\frac{2 x-1}{2 y-1}=k$

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) $\frac{d^{2} y}{d x^{2}}+(a-b) y=0$

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) $y=x e^{-x}$

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

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

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) $2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$

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) $y=A \cos x+B \sin x$

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) $x \sec x=\tan y+C$

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) $x(y+\cos x)=\cos x+C$

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) $y=\log \left\{\frac{e^{x}+1}{y+1}\right\}+k$

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) $e^{x}-e^{y}=\frac{x^{3}}{3}+c$

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) $y\left(1+x^{2}\right)=C+\sin ^{-1} x$

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

(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}

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.
$\\a.\;\; \frac{dy}{dx}+my=0\\\\ b.\;\; \frac{dy}{dx}-my=0\\\\ c.\;\; \frac{d^{2}y}{dx^{2}}-m^{2}y=0\\\\ d.\;\; \frac{d^{2}y}{dx^{2}}+m^{2}y=0\\$

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
$\\(a)\frac{\sin x}{\sin y}=c\\ (b)\sin x \sin y = c\\ (c)\sin x +\sin y = c\\ (d)\cos x \cos y = c$

$\\\cos x \sin y dx +\sin x \cos y dy=0\\ \Rightarrow \sin x \cos y dy=-\cos x \sin y dx\\ \Rightarrow \frac{\cos y}{sin y}dy=-\frac{\cos x}{sin x}dx\\ \Rightarrow \cot y dy =-\cot x dx$
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

#### Also Check NCERT Books and NCERT Syllabus here

1. Are these solutions helpful in competitive exams?

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

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

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

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

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

4. Who prepare these solutions to the maths chapter?

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

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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.

Good Luck

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.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects Â and Â 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.

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.

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

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) Option 2) Option 3) Option 4)

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 3.8×107 J 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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

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

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

With increase of temperature, which of these changes?

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

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

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

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

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

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##### Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

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##### Linguist

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

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##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

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.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

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.

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##### 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
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Computer System Analyst

Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

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##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available