NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

Question:1

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

Answer:

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

$\Rightarrow \\ 2^{-x}-2^{-y}=d$
d is also some arbitrary constant = c In2

Question:2

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

Answer:

To find: Differential equation of all non vertical lines

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

The slope of the line cannot be $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ for the given condition because if it is so, the line will become perpendicular without any necessity.
So,
$m\neq \frac{\pi}{2},m\neq \frac{3\pi}{2}$
Differentiate the general form of equation of the 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 differential equation of all non-vertical lines.

Question:3

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

Answer:

Given:
$\frac{dy}{dx}=e^{-2y}$
$(5,0)$ is a solution of this equation
To find: Solution of the given differential equation
Rewriting the equation.
$\frac{dy}{e^{-2y}}=dx$
Integrate on both the sides,
$\int \frac{dy}{e^{-2y}}=\int dx$

$\Rightarrow \int e^{2y}dy=\int dx$
Formula:
$\int e^{ax} dx =\frac{1}{a}e^{ax}$

$\frac{e^{2y}}{2}=x+c$
Given $(5,0)$ is a solution so to get c, satisfying these values
$\frac{1}{2}=5+c\\ c=-\frac{9}{2}\\$
Hence the solution is
e^2y=2x + 9
when y=3,
e²⁽³⁾ =2x + 9
$\Rightarrow$e⁶=2x + 9
$\Rightarrow$e⁶+ 9=2x
$\Rightarrow x=\frac{e^{6}+9}{2}$

Question:4

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

Answer:

Given:
$(x^{2}-1)\frac{dy}{dx}+2xy=\frac{1}{x^{2}-1}$
To find the 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 linear differential equation Compare it with,
$\frac{dy}{dx}+p(x)y=q(x)$

$p(x)=\frac{2x}{\left (x^{2}-1 \right )}$ and $q(x)=\frac{1}{\left (x^{2}-1 \right )^{2}}\\$
Calculate the Integrating Factor,
$IF=e^{\int p(x)dx}$

$\Rightarrow IF=e^{\int \frac{2x}{x^{2}-1}dx}$

$\Rightarrow IF=e^{ln(x^{2}-1)}$

$\Rightarrow \\IF=x^{2}-1$

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

$=\int \frac{dx}{x-1}+\int \frac{dx}{x+1}=ln\left ( x+1 \right )+ln\left ( x-1 \right )$

Hence, the solution of the differential equation is given by,
$y.(IF)=\int q(x).(IF)dx$

$\Rightarrow y(x^{2}-1)=\int \frac{1}{(x^{2}-1)^{2}}(x^{2}-1)dx$

$\Rightarrow y(x^{2}-1)=\int \frac{1}{(x^{2}-1)}dx$

Formula: $\int \frac{1}{(x^{2}-1)}dx=\frac{1}{2}\log\left ( \frac{x-1}{x+1} \right )\\ \frac{1}{(x^{2}-1)}dx=\frac{1}{2}\log\left ( \frac{x-1}{x+1} \right )+c$

Question:5

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

Answer:

$\frac{dy}{dx}+2xy=y$
To find: Solution of the given differential equation
$\int \frac{1}{x}{dx}=ln x+c$

$ \int x^{n}dx=\frac{x^{n+1}}{n+1}+c$
Rewriting the given equation as,
$\frac{dy}{dx}=y\left ( 1-2x \right )$

$\Rightarrow \frac{dy}{y}=\left ( 1-2x \right )dx$
Integrate on both the sides,
$\int \frac{dy}{y}=\int \left ( 1-2x \right )dx$

$\Rightarrow \ln y =\left ( x-x^{2} \right )+\log c$

$\Rightarrow \ln y-\ln c=x-x^{2}$

$\Rightarrow \ln \frac{y}{c}=x-x^{2}$

$\Rightarrow \frac{y}{c}=e^{x-x^{2}}$

$\Rightarrow y=ce^{x-x^{2}}\\$

Question:6

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

Answer:

Given:
$\frac{dy}{dx}+ay=e^{mx}\\$
It is a first order differential equation. Comparing it with,
$\frac{dy}{dx}+p(x)y=q(x)\\$
$P(x) =a$
$Q(x)=e^{xm}$
Calculating Integrating Factor
$IF=e^{\int p(x)dx}$

$IF=e^{\int a \;dx}$

$IF=e^{ax}$
Hence the solution of the given differential equation is ,
$y\left (IF \right )=\int q(x).(IF)dx$

$\Rightarrow y.(e^{ax})=\int e^{mx}e^{ax} dx$

$\Rightarrow y.(e^{ax})=\int e^{\left (m+a \right )x} dx$

$\Rightarrow y.(e^{ax})=\frac{\left (e^{(m+a)x} \right )}{m+a}+c$

Question:7

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

Answer:

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

formula:$ \int e^{x}dx=e^{x}\\ x=-e^{-t}+c$
Substituting $\; t=x+y$

$x=-e^{-(x+y)}+c$ is the solution of the differential equation.

Question:8

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

Answer:

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

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

$\Rightarrow \ln x-\frac{x^{2}}{2}+\ln c=\ln y$

$\Rightarrow -\frac{x^{2}}{2}=\ln y-\ln x-\ln c$

$\Rightarrow -\frac{x^{2}}{2}=\ln \frac{y}{cx}$

$\Rightarrow y=cxe^{-\frac{x^{2}}{2}}$

Question:9

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

Answer:

Given:
$\frac{dy}{dx}=\left ( 1+x \right )\left ( 1+y^{2} \right )$ and (0,0) is solution of the equation
To find: solution of the differential equation
Rewriting the given equation as,
$\frac{dy}{1+y^{2}}=\left ( 1+x \right )dx$
Integrating on both the sides
$\int \frac{dy}{1+y^{2}}=\int \left ( 1+x \right )dx$

$ \tan^{-1}y=x+\frac{x^{2}}{2}+c$
Substitute(0,0) to find c’s value
$0+0=c$
$c=0$
Hence, the solution is
$\tan^{-1}y=x+\frac{x^{2}}{2}$

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

Question:10

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

Answer:

Given:
$\left ( x+2y^{3} \right )\frac{dy}{dx}=y$
To find: Solution of the differential equation
Rewriting the equation as
$\frac{dx}{dy}=\frac{\left ( x+2y^{3} \right )}{y}$

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

$\Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y^{2}\\$
It is a first order linear differential equation
Comparing it with
$\frac{dx}{dy}+p(y)x=q(y)$

$ p(y)=-\frac{1}{y}$ and $q(y)=2y^{2}$
Calculation the integrating factor,
$IF=e^{\int p(y)dy}$

$\Rightarrow IF=e^{\int -\frac{1}{y}dy}$

$\Rightarrow IF=e^{-\ln y}=\frac{1}{y}$

Therefore, the solution of the differential equation is
$x.\left ( IF \right )=\int q(y).(IF)dy$

$\Rightarrow \frac{x}{y}=\int 2y \; dy$

$\Rightarrow \frac{x}{y}=y^{2}+c$

$\Rightarrow x=y^{3}+cy$

Question:11

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

Answer:

Given:

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

To find: Solution of the differential equation

Rewriting the given equation as,

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

Integrating on both sides,

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

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

Let $\sin x=t$ and $\cos xdx= dt$
$\Rightarrow \ln (1+\mathrm{y})=\int \frac{-\mathrm{dt}}{2+\mathrm{t}}$

$\Rightarrow \ln(1+y)=-\ln(2+t)+\log c$

$\Rightarrow \ln(1+y)+\ln(2+\sin x)=\log c$

$\Rightarrow(1+y)(2+\sin x)=c$

When $x=0$ and $y=1$

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

$c=4$

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

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

Question:12

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

Answer:

$(1+t)\frac{dy}{dt}-ty=1$ and $(0,-1)$ is a solution
To find: Solution for the differential equation
Rewriting the given equation as,
$\frac{dy}{dt}-\frac{ty}{1+t}=\frac{1}{1+t}$
It is a first order linear differential equation
Comparing it with,
$\frac{dy}{dt}-p(t)y=q(t)$

$p(t)=-\frac{t}{1+t}$ and $q(t)=\frac{1}{1+t}$
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,
$y.(IF)=\int q(t).(IF)dt$

$\Rightarrow y(e^{-t}(1+t))=\int e^{-t}(1+t)\frac{1}{1+t}dt$

$\Rightarrow y(e^{-t}(1+t))=\int e^{-t}dt$

$\Rightarrow y(e^{-t}(1+t))=-e^{-t}+c$
Substitution (0,-1) to find the value of c
$-1=-1+c$

$c=0$

$y(e^{-t}(1+t))=-e^{-t}$
The solution therefore y(1) is
$y(1)=-\frac{1}{1+1}=-\frac{1}{2}$

Question:13

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

Answer:

Given:

$y=(\sin^{-1} x)^{2}+A \cos^{-1} x+B$
To find: Solution of the differential equation
Differentiating on both the sides,
$\frac{dy}{dx}=2 \sin^{-1} x\frac{1}{\sqrt{1-x^{2}}}-A\frac{1}{\sqrt{1-x^{2}}}$

$\Rightarrow \sqrt{1-x^{2}}\frac{dy}{dx}=2 \sin^{-1} x-A$

Differntiate again on both the sides
$\sqrt{1-x^{2}}\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}\frac{x}{\sqrt{1-x^{2}}}=2\frac{1}{\sqrt{1-x^{2}}}$

$\Rightarrow \left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=2$
Hence the solution is
$\left ( 1-x^{2} \right )\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}-2=0\\$

Question:14

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

Answer:

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

$k=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$
Substituting the value of k in (i)
$x^{2}+y^{2}-2y\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}=0$

$\Rightarrow \left ( x^{2}+y^{2} \right )\frac{dy}{dx}-2yx+2y^{2}\frac{dy}{dx}=0$

$\Rightarrow \left ( x^{2}-y^{2} \right )\frac{dy}{dx}-2yx=0$

Question:15

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

Answer:

Given:
$\left ( 1+x^{2} \right )\frac{dy}{dx}+2xy=4x^{2}$ and $(0,0)$ is a solution to the curve
To find: Equation of the curve satisfying the differential equation
Rewrite the given equation
$\frac{dy}{dx}+\frac{2xy}{\left ( 1+x^{2} \right )}=\frac{4x^{2}}{1+x^{2}}$
Comparing with
$\frac{dy}{dx}+p(x)y=q(x)$

$p(x)=\frac{2x}{1+x^{2}}$ and $q(x)=\frac{4x^{2}}{1+x^{2}}\\$
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
$1+x^{2}=t$

$\Rightarrow 2x\; dx=dt$

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
$y.(IF)=\int q(x).(IF)dx$

$y(1+x^{2})=\int \frac{4x^{2}}{1+x^{2}}(1+x^{2})dx$

$y(1+x^{2})=\frac{4}{3}x^{3}+c$
Satisfying $(0,0)$ in the equation of the curve to find the value of c
0+0=c
c=0
Therefore equation of the curve is
$y(1+x^{2})=\frac{4}{3}x^{3}\\ y=\frac{4x^{2}}{3(1+x^{2})}$

Question:16

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

Answer:

Given:
$x^{2}\frac{dy}{dx}=x^{2}+y^{2}+xy$
To find: solution for the differential equation
Rewriting the given equation as
$\frac{dy}{dx}=1+\frac{y^{2}}{x^{2}}+\frac{y}{x}$
Clearly it is a homogenous equation
Assume $y=Vx$
Differentiate on both sides
$\frac{dy}{dx}=V+x\frac{dV}{dx}$
Substituting dy/dx in the equation
$1+\frac{y^{2}}{x^{2}}+\frac{y}{x}=V+x\frac{dV}{dx}$

$\Rightarrow1+V^{2}+V=V+x\frac{dV}{dx}$

$\Rightarrow1+V^{2}=x\frac{dV}{dx}$

$\Rightarrow \frac{dx}{x}=\frac{dV}{1+V^{2}}$
Integrating on both the sides
$\int \frac{dx}{x}=\int \frac{dV}{1+V^{2}}$

$\Rightarrow \ln x=\tan^{-1} V+c$

Substitute $V=\frac{y}{x}$

$ \ln x=\tan^{-1}\frac{y}{x}+c$
is the solution for the differential equation.

Question:17

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

Answer:

Given
$(1+y^{2})+(x-e^{\tan^{-1} y})\frac{dy}{dx}=0$
To find: Solution of the given differential equation
Rewrite the given equation as,
$(1+y^{2})\frac{dx}{dy}+x-e^{\tan^{-1} y}=0$

$\Rightarrow \frac{dx}{dy}+\frac{x}{(1+y^{2})}=\frac{e^{\tan^{-1} y}}{(1+y^{2})}$
It is a first order differential equation
Comparing it with
$\frac{dx}{dy}+p(y)x=q(y)$

$p(y)=\frac{1}{(1+y^{2})}$ and $q(y)=\frac{e^{\tan^{-1} y}}{(1+y^{2})}$
Calculating Integrating Factor
$IF=e^{\int p(y)dy}$

$IF=e^{\int \frac{1}{1+y^{2}}dy}$

$IF=e^{\tan^{-1} y}$
Hence the solution of the given differential equation is
$x.(IF)=\int q(y).(IF)dy$

$\Rightarrow x.(e^{\tan^{-1} y})=\int \frac{e^{\tan^{-1} y}}{(1+y^{2})}(e^{\tan^{-1} y })dy$

Assume $ (e^{\tan^{-1} y })=t$
Differentiate on both the sides
$\frac{e^{\tan^{-1} y }}{(1+y^{2})}dy=dt$

$x.t = \int tdt$

$x.t = \frac{t^2}{2}+c \\$

Substituting$ \; t$

$x.(e^{\tan^{-1}y})=\frac{e^{2\tan^{-1}y}}{2}+c$

Question:18

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

Answer:

Given:
$y^{2}dx+(x^{2}-xy+y^{2})dy=0$
To find: Solution for the given differential equation
Rewrite the given equation
$y^{2}\frac{dx}{dy}=xy-x^{2}-y^{2}$

$\frac{dx}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}$
It is a homogenous differential equation
Assume $x=vy$
Differentiating on both the sides
$\frac{dx}{dy}=v+y\frac{dv}{dy}$
Substitute dy/dx in the given equation
$v+y\frac{dv}{dy}=\frac{x}{y}-1-\frac{x^{2}}{y^{2}}$
Substitute v=x/y
$v+y\frac{dv}{dy}=v-1-v^{2}$

$y\frac{dv}{dy}=-1-v^{2}$

$ \frac{dv}{1+v^{2}}=-\frac{dy}{y}$
Integrating on both the sides
$\int \frac{dv}{1+v^{2}}=-\int \frac{dy}{y}$

$ \tan^{-1}v=-\ln\;y+c$

Substituting $v=\frac{x}{y}$

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

Question:19

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

Answer:

Given:
$(x+y)(dx-dy)=dx+dy$
To find: Solution of the given differential equation
Rewriting the given equation
$\left ( x+y \right )\left ( 1-\frac{dy}{dx} \right )=\left ( 1+\frac{dy}{dx} \right )$
Assume $x+y=z$
Differentiate on both sides with respect to x
$1+\frac{dy}{dx}=\frac{dz}{dx}$
Substituting the values in the equation
$z\left ( 1-\frac{dz}{dx}+1 \right )=\frac{dz}{dx}$

$\Rightarrow 2z-z\frac{dz}{dx}-\frac{dz}{dx}=0$

$\Rightarrow 2z=\left ( z+1 \right )\frac{dz}{dx}$

$\Rightarrow dx=\left ( \frac{1}{2}+\frac{1}{2z} \right )dz$
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$
Substitute $v=xy$
$x=\frac{x+y}{2}+\frac{1}{2}\ln(x+y)+\ln c$

$\Rightarrow x-y-\ln(x+y)-\ln c=0$

$\Rightarrow \ln(x+y)+\ln c=x-y$

$\Rightarrow \ln c(x+y)=x-y$

$\Rightarrow c(x+y)=e^{x-y}$

$\Rightarrow x+y=1/c(e^{x-y})$

$\Rightarrow x+y=d(e^{x-y})$ where $d=\frac{1}{c}\\$

Question:20

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

Answer:

Given:
$2\left ( y+3 \right )-xy\frac{dy}{dx}=0$
To Find: Solution of the differential equation
$2\frac{dx}{x}=\frac{ydy}{y+3}$

$\Rightarrow 2\frac{dx}{x}=\frac{\left [ \left ( y+3 \right ) -3\right ]dy}{y+3}$

$\Rightarrow 2\frac{dx}{x}=dy-\frac{3dy}{y+3}$
Integrating on both sides
$\int 2\frac{dx}{x}=\int dy-\int \frac{3dy}{y+3}$

$\Rightarrow 2\ln x=y-3\ln(y+3)+c$
Substitute $(-2,1)$ to find value of c
$\\0=-2+c$

$\Rightarrow c=2$

$2 \ln x =y-3 \ln(y+3)+2$

$\Rightarrow 2 \ln x+3\ln(y+3)=y+2$

$\Rightarrow 2 \ln x+3\ln(y+3)=y+2$

$\Rightarrow \ln x^{2}+ \ln(y+3)^{3}=y+2$

$\Rightarrow x^{2}(y+3)^{3}=e^{y+2}$

Question:21

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

Answer:

Given:
$dy=\cos x(2-y\; cosec \;x)dx$
$\left ( \frac{\pi}{2},2 \right )$ is a solution of the given differential equation
Rewriting the given equation
$\frac{dy}{dx}=2\cos x-y \cot x$

$\Rightarrow \frac{dy}{dx}+y \cot x=2\cos x\\$
It is a first order differential equation
$p(x)=\cot x$ and $q(x)=2 \cos x$
Calculate integrating factor
$IF=e^{\int p(x)dx}$

$\Rightarrow IF=e^{\int \cot x dx}$

$\Rightarrow IF=e^{\ln \sin x}$

$\Rightarrow IF=\sin x$
Therefore, the solution of the differential equation is
$y.(IF)=\int q(x).(IF)dx$

$\Rightarrow y \sin x= 2\int \cos x \sin x dx$

$\Rightarrow y \sin x =\int \sin 2x \;dx$

$\Rightarrow y \sin x = -\frac{1}{2}\cos 2x +c$
Substituting $\left ( \frac{\pi}{2},2 \right )$to find the value of c
$2= \frac{1}{2}+c$

$\Rightarrow c=\frac{3}{2}$
Hence the solution is
$y \sin x=-\frac{1}{2}\cos 2x+\frac{3}{2}$

Question:22

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

Answer:

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

Differentiate with respect to x
$2Ax+2By\frac{dy}{dx}=0$

$\Rightarrow Ax+By\frac{dy}{dx}=0....(i)$

$\Rightarrow \frac{dy}{dx}=-\frac{Ax}{By}.....(ii)$
Differentiate the curve (i) again to get,
$A+B\left ( \frac{dy}{dx}\frac{dy}{dx}+y\frac{d^{2}y}{dx^{2}} \right )=0$

$\Rightarrow -\frac{A}{B}=\left ( \left ( \frac{dy}{dx} \right )^{2}+y\frac{d^{2}y}{dx^{2}} \right )$
Substituting this in eq(i)
$\frac{dy}{dx}=-\frac{x}{y}\left ( \left ( \frac{dy}{dx} \right )^{2}+y\frac{d^{2}y}{dx^{2}} \right )$

$\Rightarrow y\frac{dy}{dx}=-x \left ( \frac{dy}{dx} \right )^{2}-xy\frac{d^{2}y}{dx^{2}}$

$\Rightarrow y\frac{dy}{dx}+x \left ( \frac{dy}{dx} \right )^{2}+xy\frac{d^{2}y}{dx^{2}}=0$

Question:23

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

Answer:

Given:
$(1+y^{2})\tan^{-1}x dx +2y(1+x^{2})dy=0$
To find: Solution for differential equation that s given
Rewriting the given equation as.
$(1+y^{2})\tan^{-1}x =-2y(1+x^{2})\frac{dy}{dx}$

$\Rightarrow \frac{\tan^{-1}x}{(1+x^{2})}dx=-\frac{2y}{(1+y^{2})}dy$
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
$\frac{1}{1+x^{2}}dx=dt$
For RHS
Assume 1+y²=z
$2ydy=dz$
Substituting and integrating on both the sides
$\int t\; dt=-\int \frac{dz}{z}$

$\Rightarrow \frac{t^{2}}{2}=-\ln z+c$
Substitute for t and z
Solution for the differential equation is
$\frac{\left ( \tan^{-1}x \right )^{2}}{2}=-\ln\left ( 1+y^{2} \right )+c$

$\Rightarrow \frac{\left ( \tan^{-1}x \right )^{2}}{2}+\ln\left ( 1+y^{2} \right )=c\\$

Question:24

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

Answer:

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

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

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

Question:25

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

Answer:

$y+\frac{dy}{dx}(xy)=x(\sin x +\log x)$
Now dx/dy (xy) refers to the differentiation of xy with respect to x
Using product rule
$ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} $
When we put it back originally in the differential equation given,
$\\ \Rightarrow y+y+x \frac{d y}{d x}=x(\sin x+\log x)$

$\Rightarrow x \frac{d y}{d x}+2 y=x(\sin x+\log x) $
Divide by $x$
$ \Rightarrow \frac{d y}{d x}+\left(\frac{2}{x}\right) y=\sin x+\log x $
Compare $\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{2}{\mathrm{x}}\right) \mathrm{y}=\sin \mathrm{x}+\log _{\mathrm{x}}$ with $ \quad \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
We get
$ \mathrm{P}=\frac{2}{\mathrm{x}} \text { and } \mathrm{Q}=\sin \mathrm{x}+\log \mathrm{x} $
The above equation is a linear differential equation with P and Q as functions of x
The first to find the solution of a linear differential equation is to find the integrating factor. $\Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{Pdx}}$
$ \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \int \frac{1}{\mathrm{x}} \mathrm{dx}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \log \mathrm{x}}$

$\Rightarrow \mathrm{IF}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2} \\ \Rightarrow \mathrm{IF}=\mathrm{x}^{2} $
The solution of the linear differential equation is
$y(\mathrm{IF})=\int Q(\mathrm{IF}) \mathrm{d} \mathrm{x}+\mathrm{c}$
Substituting values for Q and IF
$\Rightarrow y x^{2}=\int(\sin x+\log x) x^{2} d x+c$
$ \Rightarrow y x^{2}=\int x^{2} \sin x d x+\int x^{2} \log x d x+c \ldots(a) $
Find the integrals individually,
Using uv for integration
$ \\ \Rightarrow \int_{U V} d x=u \int v d x-\int\left(u^{\prime} j v\right) d x$

$\Rightarrow \int x^{2} \sin x d x=x^{2}(-\cos x)-\int 2 x(-\cos x) d x $
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 \int x \cos x d x$
$ \Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2\left(x \sin x-\int \sin x d x\right) $
$\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2(x \sin x-(-\cos x))$
$ \Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 x \sin x+2 \cos x \ldots \text { (i) } $
Now use product rule
$ \\ \Rightarrow \int x^{2} \log x d x=\log x\left(\frac{x^{3}}{3}\right)-\int\left(\frac{1}{x}\right)\left(\frac{x^{3}}{3}\right) d x$

$ \Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{1}{3} \int x^{2} d x $
$ \Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{x^{3}}{9} \ldots(i i) $
Substitute (i) and (ii) in (a)
$ \Rightarrow \mathrm{yx}^{2}=-\mathrm{x}^{2} \cos \mathrm{x}+2 \mathrm{x} \sin \mathrm{x}+2 \cos \mathrm{x}+\left(\frac{\mathrm{x}^{3}}{3}\right) \log \mathrm{x}-\frac{\mathrm{x}^{3}}{9}+\mathrm{c} $
Divide by $x^{2}$
$ \Rightarrow y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^{2}}+\left(\frac{x}{3}\right) \log x-\frac{x}{9}+\frac{c}{x^{2}} $

Question:26

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

Answer:

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

$\Rightarrow \frac{d x}{d y}+\left(\frac{2}{1+t a n y}\right) x=1$
Compare $\frac{\mathrm{dx}}{\mathrm{dy}}+\left(\frac{2}{1+\operatorname{tany}}\right) \mathrm{x}=1$ with $\quad \frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}$
We get
$\\ P=\frac{2}{1+\tan y}$ and $Q=1$
This is the linear differential equation with P and Q as functions of x
$\\\Rightarrow \mathrm{IF}=\mathrm{e}^{\int\mathrm{Pdy}}$

$\Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2}{1+\tan \mathrm{y}} \mathrm{dy}}$
Put $\tan y =\frac{\sin y}{\cos y}$
$ \Rightarrow \mathrm{IF}=\mathrm{e}^{\int \frac{2 \cos y}{\sin y+\cos y}} \mathrm{dy} $
Adding and subtracting $\sin y$ 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
$\\\frac{\mathrm{dt}}{\mathrm{dy}}=\operatorname{cosy}-\sin y$

$\Rightarrow \mathrm{dt}=(\cos \mathrm{y}-\mathrm{sin} \mathrm{y}) \mathrm{d} \mathrm{y}$
$\begin{aligned} &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\int \frac{1}{t} d t\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log t\\ &\text { Resubstitue }\\ &\Rightarrow \int \frac{\cos y-\sin y}{\sin y+\cos y} d y=\log (\sin y+\cos y)\\ &\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}+\log (\sin y+\cos y)} \end{aligned}$
$\\\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}} \times \mathrm{e}^{\text {log }(\text { siny }+\cos y)}$

$\Rightarrow \mathrm{IF}=\mathrm{e}^{\mathrm{y}}(\mathrm{sin} \mathrm{y}+\cos \mathrm{y})$
The solution of the linear differential equation will be $x(\mathrm{IF})=\int \mathrm{Q}(\mathrm{IF}) \mathrm{d} \mathrm{y}+\mathrm{c}$
Substitute values for Q and IF
$\\\Rightarrow x e^{y}(\sin y+\cos y)=\int(1) e^{y}(\sin y+\cos y) d y+c$ $\Rightarrow x e^{y}(\sin y+\cos y)=\int\left(e^{y} \sin y+e^{y} \cos y\right) d y+c$
Put $\mathrm{e}^{\mathrm{y}} \sin y=t$ and differentiate with respect to y
We get
$ \frac{d t}{d y}=e^{y} \sin y+e^{y} \cos y $
Which means
$ \mathrm{dt}=\left(\mathrm{e}^{\mathrm{y}} \sin \mathrm{y}+\mathrm{e}^{\mathrm{y}} \cos \mathrm{y}\right) \mathrm{d} \mathrm{y}+\mathrm{c} $
Hence
$ \\ \Rightarrow x e^{y}(\sin y+\cos y)=\int d t+c$

$\Rightarrow x e^{y}(\sin y+\cos y)=t+c $
Substitute t again
$x e^{y}(\sin y+\cos y)=e^{y} \sin y+c$
$ \Rightarrow \mathrm{x}=\frac{\sin \mathrm{y}}{\sin \mathrm{y}+\cos \mathrm{y}}+\frac{\mathrm{c}}{\mathrm{e}^{\mathrm{y}}(\sin \mathrm{y}+\mathrm{cosy})} $

Question:27

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

$ \frac{d y}{d x}=\cos (x+y)+\sin (x+y) $
Using the given hint substitute $x+y=z$
$ \Rightarrow \frac{\mathrm{d}(\mathrm{z}-\mathrm{x})}{\mathrm{dx}}=\cos \mathrm{z}+\sin \mathrm{z} $
Differentiate $z- x$ with respect to $x$
$ \\ \Rightarrow \frac{d z}{d x}-1=\cos z+\sin z$

$\Rightarrow \frac{d z}{d x}=1+\cos z+\sin z$

$\Rightarrow \frac{d z}{1+\cos z+\sin z}=d x $
Integrate
$ \Rightarrow \int \frac{\mathrm{d} z}{1+\cos z+\sin z}=\int \mathrm{d} \mathrm{x} $
As we know
$\cos 2 z=2 \cos ^{2} z-1$
And $\sin 2 z=2 \sin z \cos z$
$ \\ \Rightarrow \int \frac{d z}{1+2 \cos ^{2} \frac{z}{2}-1+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x$

$\Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}+2 \sin \frac{z}{2} \cos \frac{z}{2}}=x$

$\Rightarrow \int \frac{d z}{2 \cos \frac{z}{2}\left(\cos \frac{z}{2}+\sin \frac{z}{2}\right)}=x$

$\Rightarrow \int \frac{d z}{2 \cos ^{2} \frac{z}{2}\left(1+\frac{\sin \frac{z}{2}}{\cos \frac{2}{2}}\right)}=x $
$\int \frac{\sec^2 \frac{z}{2}}{2(1+\tan \frac{z}{2})}dz$
$1+\tan \frac{z}{2}=t$
Differentiate with respect to z
We get
$ \frac{d t}{d z}=\frac{\sec ^{2} \frac{z}{2}}{2} $
Hence $\frac{\sec ^{2} \frac{z}{2} \mathrm{dz}}{2}=\mathrm{dt}$
$\Rightarrow \int \frac{d t}{t}=x$
$\log t +c=x$
Again substitute t
$ \Rightarrow \log \left(1+\tan \frac{z}{2}\right)+c=x $
Similarly substitute z
$ \Rightarrow \log \left(1+\tan \frac{x+y}{2}\right)+c=x $

Question:28

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

Answer:

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

$\Rightarrow 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 $\int 3 \mathrm{e}^{-3 \mathrm{x} \frac{\cos 2 \mathrm{x}}{2} }\text { we get }$
$\Rightarrow 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} $
$\Rightarrow 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} $
$\Rightarrow\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} $
$\Rightarrow 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
$\\ \text { ye }^{- 3 x}=\int e^{-3 x} \sin 2 x d x$

$\Rightarrow \text { ye }^{-3 x}=\frac{-e^{-3 x}(2 \cos 2 x+3 \sin 2 x)}{13}+c$

$ \Rightarrow y=-\frac{1}{13}(3 \sin 2 x+2 \cos 2 x)+c e^{3 x}$

Question:29

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

Answer:

Slope of the tangent is given by $\frac{x^{2}+y^{2}}{2 x y}$
Slope of the tangent of the curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{d} \mathrm{x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{xy}}$
Put $y=vx$
$\Rightarrow \frac{\mathrm{d}(\mathrm{vx})}{\mathrm{dx}}=\frac{\mathrm{x}^{2}+\mathrm{v}^{2} \mathrm{x}^{2}}{2 \mathrm{vx}^{2}}$
Using product rule differentiate $vx$
$\\\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}=\frac{1+\mathrm{v}^{2}}{2 \mathrm{v}}$

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

$\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}-2 \mathrm{v}^{2}}{2 \mathrm{v}}$ $\Rightarrow \frac{d v}{d x}=\left(\frac{1}{x}\right) \frac{1-v^{2}}{2 v}$
$ \Rightarrow \frac{2 v}{1-v^{2}} d v=\left(\frac{1}{x}\right) d x $
Integrate
$ \Rightarrow \int \frac{2 v d v}{1-v^{2}}=\int\left(\frac{1}{x}\right) d x $
Put $1-\mathrm{v}^{2}=\mathrm{t}$
$2 \mathrm{vdv}=-\mathrm{dt}$
$ \\ \Rightarrow \int \frac{-d t}{t}=\log x+c \\ \Rightarrow-\log t=\log x+c $
Resubstitute 1
$ \Rightarrow-\log \left(1-v^{2}\right)=\log x+c $
Resubstitute v
$ \Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x+c \ldots(a) $
The curve is passing through (2,1)
Hence (2,1) will satisfy the equation (a)
Put x=1 and y=2 in (a)
$\\\Rightarrow-\log \left(1-\frac{1^{2}}{2^{2}}\right)=\log 2+c$

$\Rightarrow-\log \left(\frac{4-1}{4}\right)=\log 2+\mathrm{c}$

$\Rightarrow-\log \left(\frac{3}{4}\right)=\log 2+\mathrm{c}$

$\Rightarrow-\log \left(\frac{3}{4}\right)-\log 2=c$

$\Rightarrow-\left(\log \left(\frac{3}{4}\right)+\log 2\right)=c$
Use loga+logb=logab
$\Rightarrow-\log \frac{3}{2}=c$
Put c in equation (a)
$\\ \Rightarrow-\log \left(1-\frac{y^{2}}{x^{2}}\right)=\log x-\log \frac{3}{2}$

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

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

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

$\Rightarrow \quad \frac{3 x}{2\left(x^{2}-y^{2}\right)}=e^{0}$

$\Rightarrow 3 x=2 x^{2}-2 y^{2}$
$\begin{aligned} &\Rightarrow 2 y^{2}=2 x^{2}-3 x\\ &\Rightarrow \mathrm{y}=\sqrt{\frac{2 \mathrm{x}^{2}-3 \mathrm{x}}{2}}\\ &\text {Hence the equation of the curve is}&y=\sqrt{\frac{2 x^{2}-3 x}{2}}\\ \end{aligned}$

Question:30

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

Given: Slope of the tangent is $\frac{y-1}{x^{2}+x}$
Slope of tangent of a curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\mathrm{dy} / \mathrm{dx}$
$ \\ \Rightarrow \frac{d y}{d x}=\frac{y-1}{x^{2}+x} \\ \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^{2}+x} $
Integrate
$\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}-1}=\int \frac{\mathrm{dx}}{\mathrm{x}(\mathrm{x}+1)} \ldots(\mathrm{a})$
Use partial fraction for
$\\\Rightarrow \frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$

$\Rightarrow \frac{1}{x(x+1)}=\frac{A(x+1)+B x}{x(x+1)}$
Equate the numerator
$A(x+1)+B x=1$
Put $x=0$
$A=1$
Put $x=-1$
$B=-1$
Hence

$\Rightarrow \frac{1}{x(x+1)}=\frac{1}{x}+\frac{-1}{x+1}$
Hence equation (a) becomes,
$ \\ \Rightarrow \int \frac{d y}{y-1}=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) \mathrm{dx}$

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

Question:31

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

Answer:

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

Question:32

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

Answer:

Points on the y axis and x axis are namely $A(0,a), B(b,0)$. The midpoint of AB is $P(x,y)$.
The x coordinate of the points is given by the addition of the x coordinates of A and B divided by 2.
$\begin{aligned} &\Rightarrow x=\frac{b+0}{2}\\ &b=2 x\\ &\text { Similarly, for y coordinate }\\ &\Rightarrow y=\frac{0+a}{2}\\ &a=2 y \end{aligned}$
Therefore, the coordinates of A and B are (0,2y) and (2x,0) respectively.
AB is the tangent to curve where P is the point of contact.
Slope of the line given with two points $\left(x_{1,} y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ on it is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Here $\left(x_{1}, y_{1}\right)\left(x_{2}, y_{2}\right)$ are $(0,2 y)(2 x, 0)$ respectively.
Slope of the tangent AB is
$\frac{0-2 y}{2 x-0}$
Hence the slope of the tangent is -y/x
Slope of the tangent curve is given by,
$\\\frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=-\frac{\mathrm{dx}}{\mathrm{x}}$
Integrate
$\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}}=-\int \frac{\mathrm{dx}}{\mathrm{x}}$
$\log y =-\log x+c$

$\log y+ \log x=c$
Using $\log a+\log b=\log ab$.
$\log x y=c$
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 $(a)$
$\\\log 1=c$

$\Rightarrow c=0$
put c back in (a)
$\\\log x y=0$

$\Rightarrow x y=e^{0}$
$\Rightarrow x y=1 $
Hence the equation of the curve is $x y=1$

Question:33

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

Answer:

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

Question:34

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

Answer:

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

Question:35

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

Answer:

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

Question:36

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

Answer:

The differential equation is
$\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}=0$
Order is defined as the number which represents the highest derivative in a differential equation.
$ \frac{d^{2} y}{d x^{2}} $ is the highest derivative in the given equation is second order.

Hence the degree of the equation is 2.
Integer powers on the differentials,
$ \\ \Rightarrow\left(\frac{d y}{d x}\right)^{\frac{1}{4}}=-\frac{d^{2} y}{d x^{2}}-x^{\frac{1}{5}} \\ \Rightarrow\left(\frac{d y}{d x}\right)^{\frac{1}{4}}=-\left(\frac{d^{2} y}{d x^{2}}+x^{\frac{1}{5}}\right) $
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here differentials 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
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 the 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

$\\A. \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\\B.\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

$\\C.\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0$

$\\D.\frac{d^{2} y}{d x^{2}}+2 y=0$

Answer:

If $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is a solution of a 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$

$ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-\left(\frac{d y}{d x}+y\right)-y $
$\\ \Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}-\mathrm{y}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-2 \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0$

Question:38

The differential equation for $\mathrm{y}=\mathrm{A} \cos \alpha \mathrm{x}+\mathrm{B} \sin \alpha \mathrm{x},$ where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants is

$\\A. \frac{d^{2} y}{d x^{2}}-\alpha^{2} y=0$

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

$\\C.\frac{d^{2} y}{d x^{2}}+\alpha y=0$

$\\D.\frac{d^{2} y}{d x^{2}}-\alpha y=0$

Answer:

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

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2}(A \cos \alpha x+B \sin \alpha x) $
$\\ \text { But } y=A \cos a x+B \sin a x$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\alpha^{2} y$

$\Rightarrow \frac{d^{2} y}{d x^{2}}+\alpha^{2} y=0$
Option B is correct.

Question:39

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

Answer:

$\begin{aligned} &\text {Lets solve the differential equation }\\ &\begin{array}{l} x d y-y d x=0 \\ x d y=y d x \\ \Rightarrow \frac{d y}{y}=\frac{d x}{x} \\ \log y=\log x+c \\ \log x-\log y=c \\ \text { Using } \log a-\log b=\log a / b\end{array} \end{aligned}$

$\Rightarrow \log \frac{y}{x}=c \\ \Rightarrow \frac{y}{x}=e^{c} \\ y=xe^{c}$
$\mathrm{e}^{\mathrm{c}}$ 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 the straight line and (0,0) satisfies the equation.
Option C is correct.

Question:40

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

Answer:

Differential equation is
$ \\ \cos x \frac{d y}{d x}+y \sin x=1$

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

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

$\Rightarrow \sin x d x=-d t$
Resubstitute the value of t
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}}$
$\\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}}$

$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log {\sec \mathrm{x}}} = \sec x$
Hence IF is sec x
Option C is correct.

Question:41

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

Answer:

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

Question:42

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

Answer:

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

Question:43

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

Answer:

Given differential equation
$ \Rightarrow x \frac{d y}{d x}-y=x^{4}-3 x $
Divide though by x
$ \\ \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x^{3}-3$

$\Rightarrow \frac{d y}{d x}+\left(-\frac{1}{x}\right) y=x^{3}-3 $
Compare
$ \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(-\frac{1}{x}\right) \mathrm{y}=\mathrm{x}^{3}-3$ and $\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
$ \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}} \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{-\log \mathrm{x}} $
$\begin{aligned} &\Rightarrow e^{\int P d x}=e^{\log x^{-1}}\\ &\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log _{\mathrm{x}}^{1}} \\ &\text { Hence the IF integrating factor is}&\Rightarrow\mathrm{e}^{\int \mathrm{Pdx}}=\frac{1}{\mathrm{x}}\end{aligned}$
Option C is correct.

Question:44

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

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

Answer:

$\\ \frac{d y}{d x}-y=1$

$\Rightarrow \frac{d y}{d x}=1+y$

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

$\Rightarrow \log (1+x)=x+c$

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)
$\\\log (1+y)=x+\log 2$

$\Rightarrow\log (1+y)-\log 2=x$
Using $\log a-\log b=\log a / b$
$\\\Rightarrow \log \frac{1+y}{2}=x$

$\Rightarrow \frac{1+y}{2}=e^{x}$

$\Rightarrow1+y=2 e^{x}$

$\Rightarrow y=2 e^{x}-1$
Hence solution of differential equation is $\mathrm{y}=2 \mathrm{e}^{\mathrm{x}}-1$
Option D is correct.

Question:45

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

Answer:

$\begin{aligned} &\begin{array}{l} \frac{d y}{d x}=\frac{y+1}{x-1} \\ \Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1} \end{array}\\ &\text { Integrate }\\ &\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}+1}=\int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}-1} \end{aligned}$
$\\\log (y+1)=\log (x-1)-\log c$

$\Rightarrow \log (y+1)+\log c=\log (x-1)$
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)
$\\\Rightarrow \frac{1-1}{2+1}=c$

$\Rightarrow c=0$

$\Rightarrow \frac{x-1}{y+1}=0$

$\Rightarrow x-1=0$
So only one solution exists.
Option B is correct.

Question:46

Which of the following is a second order differential equation?

$\\A. \left(y^{\prime}\right)^{2}+x=y^{2}$

$\\B. y^{\prime} y^{\prime \prime}+y=\sin x$

$\\C. y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^{2}+y=0$

$\\D. y^{\prime}=y^{2}$

Answer:

Order is defined as the number which defines the highest derivative in a differential equation
Second order means the order should be 2 which means the highest
derivative in the equation should be $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

Let's examine each of the option given
A. $\left(y^{\prime}\right)^{2}+x=y^{2}$
The highest order derivative is $y^{\prime}$ is in first order.
B. $y^{\prime} y^{\prime \prime}+y=\sin x$
The highest order derivative is $y^{\prime \prime}$ is in second order
C. $y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^{2}+y=0$
The highest order derivative is $y^{\prime \prime \prime}$ is in third order
D. $y^{\prime}=y^{2}$
The highest order derivative is $y '$ is in first order
Option B is correct.

Question:47

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

$\left(1-x^{2}\right) \frac{d y}{d x}-x y=1$
Divide through by $\left(1-\mathrm{x}^{2}\right)$
$ \\ \Rightarrow \frac{d y}{d x}-\frac{x y}{1-x^{2}}=\frac{1}{1-x^{2}}$

$\Rightarrow \frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}} $
Compare
$\frac{d y}{d x}+\left(\frac{-x}{1-x^{2}}\right) y=\frac{1}{1-x^{2}} \quad\ and \ \ \frac{d y}{d x}+P y=Q$
We get
$ \mathrm{P}=\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}, \quad \mathrm{Q}=\frac{1}{1-\mathrm{x}^{2}} $
The IF factor is given by
$ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{-\mathrm{x}}{1-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}} $
Substitute $1-x^{2}=t$ hence
$\frac{d t}{d x}=-2 x$
Which means
$\\-x d x=\frac{d t}{2}$

$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{\mathrm{dt}}{2 \mathrm{t}}}$

$\Rightarrow e^{\int \mathrm{Pdx}}=e^{\frac{1}{2} \int \frac{\mathrm{dt}}{t}}$ $\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\frac{1}{2} \log t}$ $\Rightarrow e^{\int P d x}=e^{\log t^{\frac{1}{2}}}$

$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\mathrm{e}^{\log \sqrt{t}}$

$\Rightarrow \mathrm{e}^{\int \mathrm{P} \mathrm{d} \mathrm{x}}=\sqrt{\mathrm{t}}$
Resubstitute
$\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\sqrt{1-\mathrm{x}^{2}}$
Hence the IF integrating factor is $\sqrt{1-\mathrm{x}^{2}}$
Option C is correct.

Question:48

$\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}$ is the general solution of the differential equation:

A. $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$
B. $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\mathrm{x}^{2}}{1+\mathrm{y}^{2}}$
C. $\left(1+x^{2}\right) d y+\left(1+y^{2}\right) d x=0$
D. $\left(1+x^{2}\right) d x+\left(1+y^{2}\right) d y=0$

Answer:

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

Question:49

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

Answer:

$\\ y \frac{d y}{d x}+x=c$

$\Rightarrow y \frac{d y}{d x}=c-x$

$\Rightarrow y d y=(c-x) d x$
Integrate
$\\y d y=(c-x) d x$

$\Rightarrow \int y d y=\int(c-x) d x$

$\Rightarrow \int y d y=\int c d x-\int x d x$

$\Rightarrow \frac{y^{2}}{2}=c x-\frac{x^{2}}{2}+k$
k is the integration constant
$\\\Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}=c x+k$

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

Question:50

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

Answer:

$\\e^{x} \cos y d x-e^{x} \sin y d y=0$

$\Rightarrow e^{x} \cos y d x=e^{x} \sin y d y$

$\Rightarrow d x=\frac{\sin y}{\cos y} d y$
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 $\sin y dy=-dt$
$\\\quad \Rightarrow x=\int \frac{-d t}{t}$

$x=-\log t+c$

$x+c=\log (\cos y)^{-1}$
$\\ \Rightarrow x+c=\log \frac{1}{\cos y}$

$x+c=\log (\sec y)$

$e^{x+c}=\sec y$

$\mathrm{e}^{x } \mathrm{e}^{c}=\sec y$

$\Rightarrow \mathrm{e}^{\mathrm{x}}=\frac{1}{\mathrm{e}^{\mathrm{c}} \operatorname{cosy}}$

$e^{x} \cos y=e^{-c}$

$\mathrm{e}^{-\mathrm{c}}$ is constant because e is a constant and $\mathrm{c}$ is the integration constant let us it denote as $\mathrm{k}$ hence $\mathrm{e}^{-c}=\mathrm{k}$
$ e^{x} \cos y=k $
Option A is correct.

Question:51

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

Answer:

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

Question:52

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

Answer:

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

$\Rightarrow \quad y \cdot e^{x}=\int d x+C$

$\Rightarrow \quad y \cdot e^{x}=x+C$
Given that when x=0 and y=0
$\\\Rightarrow$ $0=0+C$

$\Rightarrow$ $C=0$
Eq. (i) becomes $y \cdot e^{x}=x$
$\Rightarrow y=x e^{-x}$

Question:53

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

Answer:

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

$ \text { I.F. }=e^{\int P d x}=e^{\int \operatorname{tan} x d x}=e^{(\log sec x)}=\sec x$

Question:54

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

Answer:

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

Question:55

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

(c) $x e^{x}$
(d) $e^{x}$

Answer:

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

$ \mathrm{IF},=\int P d x=e^{-\int \frac{1-x}{x} d x} = \frac{e^x}{x}$

Question:58

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

Answer:

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

$\Rightarrow y \cdot x=\int e^{x} d x$

$\Rightarrow y \cdot x=e^{x}+k$

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

Question:59

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

Answer:

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

$\Rightarrow a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}$
$\\$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$

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

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

Question:60

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

Answer:

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

Question:61

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

Answer:

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

Question:62

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

Answer:

The answer is the option (d) Rectangular Hyperbola
Explanation: -
According to the question, $\frac{d y}{d x}=\frac{x}{y}$

$\Rightarrow y d y=x d x$
On integrating both sides, we get
$ \frac{y^{2}}{2}=\frac{x^{2}}{2}+C $
$\Rightarrow y^{2}-x^{2}=2 C,$ which is an equation of rectangular hyperbola.

Question:63

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

Answer:

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

$\Rightarrow y e^{-x^{2} / 2}=\int 1 d x+C$

$\Rightarrow y e^{-x^{2} / 2}=x+C$

$\Rightarrow y=(x+C) e^{x^{2} / 2}$

Question:64

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

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

Question:65

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

Answer:

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

$\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0$

Question:66

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

Answer:

The answer is the option (d) $y=x e^{-x}$
Explanation: -
$\frac{d y}{d x}+y=e^{-x}, y(0)=0$
Here, $P=1$ and $Q=e^{-x}$

$I.F.=e^{\int \operatorname{ldx}}=e^{x}$
$\\\therefore$ The general solution is $y \cdot e^{x}=\int e^{-x} \cdot e^{x} d x+C$

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

$\Rightarrow$ $y e^{x}=x+C$
Given, when x=0 and y=0
$\\\Rightarrow$ $0=0+C$

$\Rightarrow C=0$
Eq. (i) reduces to $y \cdot e^{x}=x$ or $y=x e^{-x}$

Question:67

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

Answer:

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

Question:68

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

Answer:

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

Question:69

The differential equation of family of curves $y^{2}=4 a(x+a)$ is

(a) $y^{2}=4 \frac{d y}{d x}\left(x+\frac{d y}{d x}\right)$
(b) $2 y \frac{d y}{d x}=4 a$
(c) $\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0$
(d) $2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}-y=0$

Answer:

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

Question:70

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

Answer:

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

Question:71

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

Answer:

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

The general solution is $y \sec x=\int{\sec } x \cdot \sec x+C$

$\Rightarrow$ $y \sec x=\int \sec ^{2} x d x+C$

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

Question:72

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

Answer:

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

$\therefore$ The general solution is $y \cdot x=\int x \cdot \sin x d x+C$
$ \begin{aligned} &=-x \cos x-\int-\cos x d x \\ &=-x \cos x+\sin x+c \\ \Rightarrow x(y+\cos x) &=\sin x+C \end{aligned} $

Question:73

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

Answer:

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

$\Rightarrow \int\left(1-\frac{1}{y+1}\right) d y=\int \frac{e^{x}}{e^{x}+1} d x$

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

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

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

Question:74

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

Answer:

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

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

$\Rightarrow \int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x$

$\Rightarrow e^{y}=e^{x}+\frac{x^{3}}{3}+C$

$\Rightarrow e^{y}-e^{x}=\frac{x^{3}}{3}+C$

Question:75

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

Answer:

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

$\therefore$ The general solution is
$ y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \frac{1}{\left(1+x^{2}\right)^{2}}+C $
$\\\Rightarrow y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+C$

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

Question:76

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

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

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

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

Answer:

(i) Given differential equation is
$ \frac{d^{2} y}{d x^{2}}+e^{\frac{d y}{d x}}=0 $

Degree of this equation is not defined as it cannot be expresses as polynomial of derivatives.
(ii) We have $\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=x$
$\Rightarrow 1+\left(\frac{d y}{d x}\right)^{2}=x^{2}$
So, degree of this equation is two.
(iii) Given that the general solution of a differential equation has three arbitrary constants. So we require three more equations to eliminate these three constants. We can get three more equations by differentiating the 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 a linear differential equation.
(v) We have $\frac{d x}{d y}+P_{1} x=Q_{1}$
To solve such equations we multiply both sides by
So we get $e^{\int P_{1} d y}\left(\frac{d x}{d y}+P_{1} x\right)=Q_{1} e^{\int P_{1} d y}$
$\\\Rightarrow \frac{d x}{d y} e^{\int P_{1} d y}+P_{1} e^{\int P_{1}dy}=Q_{1} e^{\int P_{1}d y}$

$\Rightarrow \frac{d}{d y}\left(x e^{\int P_{1} d y}\right)=Q_{1} e^{\int P_{1} dy}$ $\Rightarrow \quad \int \frac{d}{d y}\left(x e^{\int P_{1} d y}\right) d y=\int Q_{1} e^{P_{1} d y}dy$

$\Rightarrow \quad x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$
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 \mathrm{I.F.}=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, $y d x+(x+x y) d y=0$

$\Rightarrow \quad y d x+x(1+y) d y=0$
$\Rightarrow$ $\frac{d x}{-x}=\left(\frac{1+y}{y}\right) d y$

$\Rightarrow \int \frac{1}{x} d x=-\int\left(\frac{1}{y}+1\right) d y$

$\Rightarrow \quad \log x=-\log y-y+\log C$

$\Rightarrow \quad \log x+\log y-\log C=-y$

$\Rightarrow \quad \log \frac{x y}{C}=-y$

$\Rightarrow \quad \frac{x y}{C}=e^{-y}$
$\Rightarrow \quad x y=C e^{-y}$
(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 $\cot y d x=x d y$
$\Rightarrow \int \frac{1}{x} d x=\int \tan y d y$

$\Rightarrow \log x=\log \sec y+\log C $
$\\\Rightarrow$ $ \log \frac{x}{\sec y}=\log C$

$\Rightarrow \frac{x}{\sec y}=C$

$\Rightarrow$ $x=C \sec y$
(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 a linear differential equation.
$ \therefore \text { I.F. }=e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x-\log x}=e^{x} \cdot e^{-\log x}=\frac{e^{x}}{x} $

Question:77

(i)Integrating factor of the differential of the form $\frac{d x}{d y}+P_{1} x=Q_{1}$ is given by $e^{\int P_{1}d y}$
(ii)Solution of the differential equation of the type $\frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}$ is given by $x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$
iii)Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x} f(x, y),$ where $f(x, y)$ is a homogeneous function of zero degree is y=v x
(iv)Correct substitution for the solution of the differential equation of the type $\frac{d x}{d y} g(x, y)$where g(x, y) is a homogeneous function of the degree zero is x=v y
(V)Number of arbitrary constants in the particular solution of a differential equation of order two is two.
(vi)The differential equation representing the family of circles $x^{2}+(y-a)^{2}=$ $a^{2}$ will be of order two.
(vii)The solution of $\frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3}$ is $y^{\frac{2}{3}}-x^{\frac{2}{3}}=c$
(viii)Differential equation representing the family of curves $y=e^{x}(A \cos x+$$B \sin x$ ) is $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$
ix)The solution of the differential equation $\frac{d y}{d x}=\frac{x+2 y}{x}$ is $x+y=k x^{2}$.
x)Solution of $\frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right) \text { is } \sin \frac{y}{x}=C x$
xi)The differential equation of all non horizontal lines in a plane is $\begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}$

Answer:

i) Integrating factor of the differential of the form $\frac{d x}{d y}+P_{1} x=Q_{1}$ is given by $e^{\int P_{1}d y}$.Hence given statement is true.
(ii) Solution of the differential equation of the type $\frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}$ is given by $x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C$.
Hence given statement is true.
iii) Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x} f(x, y),$ where $f(x, y)$ is a homogeneous function of zero degree is $y=v x.$
Hence given statement is true.
(iv) Correct substitution for the solution of the differential equation of the type $\frac{d x}{d y} g(x, y)$where $g(x, y)$ is a homogeneous function of the degree zero is $x=v y.$
Hence given statement is true.
(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is False.
(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 the 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.