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KC Sinha Solution Class 12 Chapter 23 अवकल समीकरण (Differential Equations) Exercise 23.7

 Exercise 23.7

Question 1

x \frac{d y}{d x}-3 y=x^{2}

Sol :

x से भाग देने पर

\frac{d y}{dx}-\frac{3 y}{x}=x

\frac{d y}{dx}+P y=Q

P=-\frac{3}{x}, Q=x


Integral factor =e^{\int pd x}

=e^{-\int \frac{3}{x} dx}=e^{-3 \log |x|}

=e^{\log \left|x^{-3}\right|}=x^{-3}=\frac{1}{x^{3}}

Integral factor=\frac{1}{x^3}

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot \frac{1}{x^{3}}=\int x \cdot \frac{1}{x^{3}} d x+C

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

y=-x2+Cx3


Question 2

x \cos x \frac{d y}{d x}+y(x \sin x+\cos x)=1

Sol :

xcos x से भाग देने पर

\frac{x \operatorname{cos} \frac{d y}{d x}}{x \cos x}+y \frac{(x \sin x+\cos x)}{x \cos x}=\frac{1}{x \cos x}

\frac{d y}{d x}+y\left(\frac{x \sin x}{x \cos x}+\frac{\cos x}{x \cos x}\right)=\frac{1}{x \cos x}

\frac{d y}{d x}+y\left(\tan x+\frac{1}{x}\right)=\frac{1}{x \cos x}

P=\tan x+\frac{1}{x}, Q=\frac{1}{x \cos x}

Integral factor=e^{\int Pd x}=e^{\int(\tan x+\frac{1}{x}) d x}

=e^{{\log} |x\sec x|+\log |x|}

=e^{\log |x \sec x|}=x sec x

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y x \sec x=\int \frac{1}{x \cos x} x \sec x d x+C

yxsec x=\int \sec ^{2} x d x+c

yxsec x=tan x+C


Question 3

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

Sol :

(1-x2) से भाग देने पर

\frac{d y}{d x}-\frac{x y}{1-x^{2}}=\frac{x}{1-x^{2}}

P=\frac{-x}{1-x^{2}},Q=\frac{x}{1-x^{2}}

Integral factor=e^{\int Pd x}=e^{-\int \frac{x}{1-x^{2}}} dx

=e^{\frac{1}{2} \int \frac{-2 x}{1-x^{2}} d x}

=e^{\frac{1}{2} \log |{1}-x^{2}|}=e^{\log \left|\sqrt{1-x^{2}}\right|}

=\sqrt{1-x^{2}}

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot \sqrt{1-x^{2}}=\int \frac{x}{1-x^{2} } \times \sqrt{1-x^{2}} d x+c

y \cdot \sqrt{1-x^{2}}=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} d x+c

y \sqrt{1-x^{2}}=-\frac{1}{2} \times 2 \sqrt{1-x^{2}}+C

y=\frac{-1 \sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}+\frac{C}{\sqrt{1-x^{2}}}

-1+\frac{C}{\sqrt{1-x^{2}}}


Question 4

\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}

Sol :

दोनो तरफ (1+x2) से भाग देने पर

\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{4 x^{2}}{1+x^{2}}

P=\frac{2 x}{1+x^{2}}, Q=\frac{4 x^{2}}{1+x^{2}}

Integral factor=e^{\int Pd x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}

Integral factor=1+x2

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot\left(1+x^{2}\right)=\int \frac{4 x^{2}}{1+x^{2}} \times \left(1+x^{2}\right) d x+C

y\left(1+x^{2}\right)=4 \cdot \frac{x^{3}}{3}+C


Question 5

x \frac{d y}{d x}+2 y=x^{2} \log x

Sol :

x से भाग देने पर

\frac{d y}{d x}+\frac{2 y}{x}=x \log x

P=\frac{2}{x},Q=x log x

Integral factor=e^{\int Pd x}=e^{\int \frac{2}{x} d m}=e^{2 \log x}

=e^{\log {x} ^2}=x^{2}

Integral factor=x2

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot x^{2}=\int x \log x \cdot x^{2} d x+c

y \cdot x^{2}=\int x \log x \cdot x^{2} d x+c

y \cdot x^{2}=\int x^{3} \cdot \log x+c

y \cdot x^{2}=\log x \int x^{3} dx-\int\left\{\frac{d(\log x)}{dx} \int x^{3} dx\right\} d x+C

y \cdot x^{2}=\frac{x^{4} \log x}{4}-\int \frac{1}{x}\times \frac{x^{4}}{4} d x+c

y \cdot x^{2}=\frac{x^{4}}{4} \log x-\frac{1}{4} \int x^{3} dx+C

y \cdot x^{2}=\frac{x^{4}}{4} \log x-\frac{1}{4} \times \frac{x^{4}}{4}+C

y x^{2}=\frac{x^{4}}{16}(4 \operatorname{log} x-1)+C

y=\frac{x^{2}}{16}(4 \log x-1)+\frac{C}{x^{2}}


Question 6

\left(x^{2}+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^{2}+4}

Sol :

(x2+1) से भाग देने पर

\frac{d y}{d x}+\frac{2 x y}{x^{2}+1}=\frac{\sqrt{x^{2}+4}}{x^{2}+1}

P=\frac{2 x}{x^{2}-1}, Q=\frac{\sqrt{x^{2}+4}}{x^{2}+1}

Integral factor$=e^{\int Pd x}

=e^{\int \frac{2 x}{x^{2}+1}dx}=e^{\log | x^{2}+1|}

=x2+1

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot\left(x^{2}+1\right)=\int \frac{\sqrt{x^{2}+4}}{x^{2}+1} \times \left(x^{2}+1\right) dx+c

\int \sqrt{x^{2}+a^{2}} dx=\frac{1}{2}x\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}|+C

y \cdot\left(x^{2}+1\right)=\int \sqrt{x^{2}+2^{2}} d x

\left.y \cdot\left(x^{2}+1\right)=\frac{1}{2} \times \sqrt{x^{2}+2^{2}}+\frac{2^{2}}{2} \log y \mid x+\sqrt{x^{2}+2^{2}}\right]+C

y\left(x^{2}+1\right)=\frac{1}{2} \times \sqrt{x^{2}+4}+2 \log \left|x+\sqrt{x^{2}+4}\right|+C


Question 7

\frac{d y}{d x}+2 y=e^{-x}

Sol :

P=2 , Q=e-x

Integral factor=e^{\int p dx}=e^{2 \int dx}=e^{2 x}

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot e^{2 x}=\int e^{-x} \cdot e^{2 x} dx+c

y \cdot e^{2 x}=\int e^{x} d x+c

y.e2x=ex+C


Question 8

\frac{d y}{d x}+2 y=6 e^{x}

Sol :

P=2 , Q=6ex

Integral factor=e^{\int Pdx}=e^{\int 2 dx}=e2x

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y e^{2 x}=\int 6 e^{x} \cdot e^{2 x} d x+C

y e^{2 x}=6 \int e^{3 x} d x+C

y e^{2 x}=6\times \frac{e^{3x}}{3}+C

y.e2x=2.e3x+C


Question 9

4 \frac{d y}{d x}+8 y=5 e^{-3 x}

Sol :

4 से भाग देने पर

\frac{d y}{d x}+2 y=\frac{5}{4} e^{-3 x}

P=2, Q=\frac{5}{4} e^{-3 x}

Integral factor=e^{\int Pdx}=e^{\int 2 dx}=e2x

अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y-e^{2 x}=\frac{5}{4} \int e^{-3 x} \cdot e^{2 x} dx+C

y e^{2 x}=\frac{5}{4} \int e^{-x} d x+C

y \cdot e^{2 x}=-\frac{5}{4} e^{-x}+C


Question 10

\frac{d y}{d x}+y=e^{-2 x}

Sol :

P=1 , Q=e-2x

Integral factor=ex

y e^{x}=\int e^{-2 x} \cdot e^{x} d x+c

y e^{x}=\int e^{-x} dx+c \Rightarrow y e^{x}=-e^{-x}+C


Question 11

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

Sol :

x से भाग देने पर

\frac{d y}{dx}=\frac{x+y}{x}

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

\frac{d y}{dx}-\frac{y}{x}=1 

P=-\frac{1}{x} , Q=1

Integral factor=e^{\int P dx}=e^{-\int \frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}

दिए गए अवकल समीकरण का हल:

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot \frac{1}{x}=\int 1 \times \frac{1}{x} d x+c

\frac{y}{x}=\log |x|+c


Question 12

\frac{d y}{d x}+3 y=e^{-2 x}

Sol :

P=3 , Q=e-2x

Integral factor=e^{\int Pdx}=e^{\int 3 d x}=e3x

दिए गए अवकल समीकरण का हल:

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y e^{3 x}=\int e^{-2 x} \cdot e^{3 x} d x+c

y e^{3 x}=\int e^{x} d x+c

ye3x=ex+C

y=e-2x+Ce-3x


Question 13

x \frac{d y}{d x}+2 y=x^{2}, x \neq 0

x से भगा देने पर

\frac{d y}{d x}+\frac{2 y}{x}=x

P=\frac{2}{x} , Q=x

Integral factor=e^{\int Pdx}=e^{\int \frac{2}{x} dx}=e^{2 \log x}

=e^{\operatorname{log} x^{2}}=x^{2}

दिए गए अवकल समीकरण का हल:

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot x^{2}=\int x \cdot x^{2} d x+C

y \cdot x^{2}=\int x^{3} d x+C

y \cdot x^{2}=\frac{x^{4}}{4}+C

y=\frac{x^{2}}{4}+\frac{C}{x^{2}}


Question 14

\frac{d y}{d x}+y=\sin x

Sol :

P=1 , Q=sin x

Integral factor

=e^{\int P dx}=e^{\int 1 d x}=ex

अवकल समीकरण का हल:

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot e^{x}=\int \sin x \cdot e^{x} d x+c

माना I=\int \sin x \cdot e^{x} d x

I=\sin \int e^{x} d x-\int\left\{\frac{d(sin x-x)}{d n} \cdot \int e^{x} d x\right\} d x

I=e^{x} \sin x-\int \cos x \cdot e^{x} dx

I=e^{x} \sin x-\left[\cos x \int e^{x} d x-\int\left\{\frac{d (\cos x)}{dx} \int e^{x} dx\right\}dx\right]

I=e^{x} \sin x-e^{x} \cos x-\int \sin x \cdot e^{x} d x

I=ex(sin x-cos x)-I

2I=ex(sin x-cos x)

I=\frac{e^{x}}{2}(\sin x-\cos x)

∴अवकल समीकरण का हल

y \cdot e^{x}=\int \operatorname{sin} x \cdot e^{x}dx+c

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

y=\frac{1}{2}(\sin x-\cos x)+Ce^{-x}


Question 15

\frac{d y}{d x}+y=\cos x

Sol :

P=1, Q=cos x


Question 16

\frac{d y}{d x}-y=\cos x

Sol :

P=1 , Q=cos x

Integral factor=e^{\int P dx}=e^{-\int 1 d x}=e-x




Question 17

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

Sol :

(x2-1) से भाग देने पर

\frac{d y}{d x}+\frac{2(x+2)}{x^{2}-1} y=\frac{2(x+1)}{x^{2}-1}

\frac{d y}{d x}+\frac{2(x+2)}{x^{2}-1} y=\frac{2(x+1)}{(x-1)(x+1)}

P=\frac{2(x+2)}{x^{2}-1}, Q=\frac{2}{x-1}

Integral factor=e^{\int P dx}=e^{-\int 1 d x}=e-x

=e^{\int \frac{2(x+2)}{x^{2}-1} d x}=e^{\int \frac{2 x}{x^{2}-1}} d x+4\int \frac{1}{x^{2}-1} d x

=e^{\log \left|x^{2}-1\right|+4 \times \frac{1}{2\times 1} \log\left|\frac{2-1}{x+1}\right|}

=e^{\log \left|x^{2} -1\right| +\log \left|\frac{(x-1)^{2}}{x+1}\right|}

=e^{\log \left|(x-1)(x+1) \cdot \frac{(x-1)^{2}}{(x+1)^{2}}\right|}

=\frac{(x-1)^{3}}{x+1}

∴अवकल समीकरण का हल

y \cdot e^{x}=\int \operatorname{sin} x \cdot e^{x}dx+c

y\frac{(x-1)^{3}}{x+1}=\int \frac{2}{x-{1}} \cdot\times \frac{(x-1)^{3}}{2+1} d x

y \frac{(x-1)^{3}}{x+1}=2 \int \frac{x^{2}-2 x+1}{x+1} dx

y \frac{(x-1)^{3}}{x+1}=2 \int\left[x-3+\frac{4}{x+1}\right] d x

y \frac{(x-1)^{3}}{x+1}=2\left[\frac{x^{2}}{2}-3 x+4 \log |x+1|\right]

y=\frac{(x+1)}{(x-1)^{3}}\left[x^{2}-6 x+8 \log (x+1)\right]+C


Question 18

\frac{d y}{d x}+y=e^{x}

Sol :
P=1 , Q=ex

Integral factor=e^{\int P dx}=e^{1 d x}=ex

दिए गए अवकल समीकरण का हल

y \cdot e^{x}=\int \sin x \cdot e^{x} d x+C

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

y e^{x}=\int e^{2 x} dx+C

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

y=\frac{1}{2} e^{x}+Ce^{-x}


Question 19

\frac{d y}{d x}-4 y=e^{2 x}

Sol :
P=-4 , Q=e2x

Integral factor$=e^{\int P dx}
=e^{-\int 4 dx}=e^{-4 x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \cdot\left(e^{-4 x}\right)=\int e^{2 x} \cdot e^{-4 x} d x+c

y e^{-4 x}=\int e^{-2 x} dx+c

\frac{y}{e^{4x}}=-\frac{1}{2} e^{-2 x}+C

y=-\frac{1}{2} e^{2 x}+Ce^{4 x}


Question 20

x \frac{d y}{d x}+3 y=x^{2}

Sol :
x से भाग देने पर

\frac{d y}{d x}+\frac{3 y}{x}=x

P=\frac{3}{x}, Q=x

Integral factor=e^{\int P dx}=e^{\int \frac{3}{x}dx}=e^{3 \log x}=x3

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y x^{3}=\int x \cdot x^{3} d x+C_1

y x^{3}=\int x^{4} d x+C_1

y x^{3}=\frac{x^{5}}{5}+C_1

y x^{3}-\frac{x^{5}}{5}=C_{1}

\frac{5 x^{3} y-x^{5}}{5}=C_1

5x3y-x5=5C1

5x3y-x5=5C , जहाँ C=5C1

Question 21

\frac{d y}{d x}+2 y=4 x



Question 22

x \frac{d y}{d x}-y=x+1
Sol :
x से भाग देने पर

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

P=-\frac{1}{x}, Q=1+\frac{1}{x}

Integral factor=e^{\int P dx}=e^{-\int \frac{1}{x} dx}=e^{-\log x}=\frac{1}{x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \frac{1}{x}=\int\left(1+\frac{1}{x}\right) \frac{1}{x} d x+c

\frac{y}{x}=\int\left(\frac{1}{x}+\frac{1}{x^2}\right) d x+c

\frac{y}{x}=\log |x|-\frac{1}{x}+c

\frac{y}{x}=\frac{x\log |x|-1+Cx}{x}

y=xlog |x|-1+Cx

Question 23

\frac{d y}{d x}+y=\cos x-\sin x
Sol :
P=1 , Q=cos x-sin x

Integral factor=e^{\int p dx}=e^{\int{1 dx}}=e^{x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y e^{x}=\int(\cos x-\sin x) e^{x} d x+C

yex=ecos x+C

y=cos x+Ce-x


Question 24

\frac{d y}{d x}+\frac{y}{x}=x^{n}
Sol :
p=\frac{1}{2}, Q=xn

Integral factor=e^{\int p d x}=e^{\int \frac{1}{x} dx}=e^{\log x}=x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot x=\int x^{n} \cdot x d x+C

y x=\int x^{n+1} d x+C

y x=\frac{x^{n+2}}{n+2}+C

x y=\frac{x^{n+2}+c(n+2)}{n+2}

(n+2)xy=xn+2+(n+2)C


Question 25

\frac{d y}{d x}-y \tan x=e^{x} \sec x
Sol :
P=-tan x , Q=esec x

Integral factor=e^{\int p d x}=e^{-\int \tan x dx}

=e^{\log (\cos x)}=cos x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \cos x=\int e^{x} \sec x \cdot \cos x dx+c

y \cos x=\int e^{x} d x+c

y cos x=ex+C

Question 26

\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\cos x
Sol :
\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cos x}{1+x^{2}}

P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{\cos x}{1+x^{2}}

Integral factor=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}}} d n

=e^{\log |1+x^2|}=1+x2

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y\left(1+x^{2}\right)=\int \frac{\cos x}{1+x^{2}} \cdot\left(1+x^{2}\right) d x+c

y(1+x2)-sin x=C

Question 27

x \frac{d y}{d x}+2 y=\sin x
Sol :
x से भाग देने पर

\frac{d y}{d x}+\frac{2 y}{x}=\frac{\sin x}{x}

P=\frac{2}{x}, Q=\frac{\sin x}{x}

Integral factor=e^{\int p d x}=e^{\int \frac{2}{x} dx}=e^{2 \log x}
=x2

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot x^{2}=\int \frac{\sin x}{x} \times x^{x} dx+C

y x^{2}=x \int \sin x dx-\int\left\{\frac{d(x)}{d x} \int \sin x dx\right\} d x+C

x^{2} y=-x \cos x+\int \cos x d x+C

x2y=-xcos x+sin x+C

Question 28

(\sec x) \frac{d y}{d x}=y+\sin x
Sol :
\frac{d y}{d x}=\frac{y}{\sec x}+\frac{\sin x}{\operatorname{sec} x}

P=- cos x, Q=sin x cos x

Integral factor=e \int {p d x}=e^{-\int \cos x dx}=e^{-\sin x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y\times e^{-\sin x}=\int \sin x \cos x \cdot e^{-\sin x} dx+c

Put sin x=t
differentiating w.r.t x
cos xdx=dt

y \cdot e^{-\sin x}=\int t e^{-t} d t+C

y \cdot e^{-\operatorname{sin} x}=t \int e^{-t} d t-\int \left\{ \frac{d(t)}{dt} \cdot \int e^{-1} dt\right\} d t+C
y e^{-\sin x}=-e^{-t} \cdot 1+\int e^{-t} d t+c

y.e-sin x=-ex-t.t+(-e-t)+C
y.e-sin x=-e-t(t+1)+C
ye-sin x=-e-sin x(sin x+1)+C
y=-(sin x+1)+Cesin x
y=Cesin x-(1+sin x)

Question 29

\frac{d y}{d x}+y \cot x=x
Sol :
P=cot x, Q=x

Integral factor=e^{\int P dx}=e^{\int \cot dx}=e^{\log |\sin x|}
=sin x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \sin x=\int x \sin x d x+C

y \sin x=x \int \sin x d x-\int\left\{\frac{d(x)}{d x} \cdot \int \sin x d x\right] d x+c

y \sin x=-x \cos x+\int \cos x dx+C

ysin x=-x cos x+sin x+C

ysin x-sin x+x cos x=C

(y-1)sin x+x cos x=C


Question 30

\frac{d y}{d x}+y \cos x=\sin x \cos x
Sol :
P=cos x, Q=sin x. cos x

Integrating factor=e^{\int P dx}

=e^{\int \cos x dx}=e^{\sin x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y e^{\sin x}=\int \sin x \cos x e^{ \sin x}d x+c

Put sin x=t
Differentiating w.r.t x
cos x dx=dt

y.esin x=\int e^{t} d t+c

y.esin x=t \int e^{t} d t-\int \left\{ \frac{d(t)}{d t} \cdot \int e^{t} d t\right\} d t+C

y.esin x=e^{t} \cdot t-\int e^{t} d t+C

y.esin x=et.t-et+C

y.esin x=et(t-1)+C

y.esin x=esin x(sin x-1)+C

y=sin x-1+Ce-sin x

Question 31

\frac{d y}{d x}+2 y \cot x=3 x^{2} \operatorname{cosec}^{2} x
Sol :
P=2 cot x , Q=3x2 cosec2 x

Integrating factor=e^{\int P dx}

=e^{2 \int \operatorname{cot x dx}}=e^{2 \log |\sin x|}
=sin2x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \sin^{2} x=\int \left(3 x^{2} \cdot \text{cosec}^{2} x \cdot \operatorname{sin}^{2} x d x+C\right.

y \sin^2 x=3\int x^{2} d x+C

y \sin ^{2} x=\frac{3x^3}{3}+C

ysin2 x=x3 +C


Question 32

x \frac{d y}{d x}-y=2 x^{2} \sec x
Sol :
x से भाग देने पर

\frac{d y}{d x}-\frac{y}{x}=2 x \sec x

P=-\frac{1}{x} , Q=2x sec x

Integrating factor=e^{\int P dx}

=e^{-\int \frac{1}{x} dx}=e^{-\log x}=\frac{1}{x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \frac{1}{x}=\int 2 x \sec x \times \frac{1}{x} d x+C

\frac{y}{x}=2 \int \operatorname{sec} x d x+C

\frac{y}{x}=2 \log |\sec x+\tan x|+C

y=2x log |sec x+tan c|+Cx

Question 33

\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x
Sol :
P=tan x, Q=2x+x2 tanx 

Integrating factor=e^{\int P dx}

=e^{\int \tan x d x}=e^{\log |\sec x|}
=sec x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \sec x=\int\left(2 x+x^{2} \tan x\right) \sec x d x+c

y \sec x=\int 2 x \sec x d x+\int x^{2} \cdot \sec x \tan x d x+C

y \sec x=\int 2 x \sec x+x^{2} \int \operatorname{sec} x \tan x d x\left.-\int\left(\frac{d\left(x^{2}\right)}{dx} \int \operatorname{sec} x.\tan x d x\right)\right] d x+C

y \sec x=\int 2 x \operatorname{sec} x+x^{2} \sec x-\int 2 x \cdot \sec x d x+C

y sec x=x2 sec x+C

Question 34

\frac{d y}{d x}+\frac{y}{x}=e^{x}
Sol :
P=\frac{1}{x} , Q=ex

Integrating factor=e^{\int P dx}

=e^{\int \frac{1}{x} dx}=e^{\log x}=x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot x=\int e^{x} \cdot x dx+C

y \cdot x=x \int e^{x} d-\int\left\{\frac{d(x)}{dx} \cdot \int e^{x} d x\right\} d x+C

y x=x \cdot e^{x}-\int 1 \cdot e^{x} d x+c

y.x=xex-ex+C

y=e^{x}-\frac{1}{x} e^{x}+\frac{C}{x}

Question 35

x \frac{d y}{d x}=y(\log y-\log x-1)
Sol :
\frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}-1\right)...(i)

यह एक समघातीय अवकल समीकरण है।
y=vx⇒v=\frac{y}{x}

Differentiating w.r.t x

\frac{d y}{d x}=\frac{d v}{dx} \cdot x+v...(ii)

समीकरण (i) तथा (ii) से,

\frac{dv}{d x} \cdot x+v=\frac{y}{x}\left(\log \frac{y}{x}-1\right)

\frac{d v}{d x} \cdot x+v=v(\log v-1)

\frac{d v}{d x} \cdot x+v=v \log v-v

\frac{dv}{dx}.x=v\log-2v

\frac{d v}{dx} \cdot x=v(\log v-2)

\frac{dv}{v(\log v-2)}=\frac{d x}{x}

Integrating both sides

\int \frac{d v}{v\left(\log v-2\right)}=\int \frac{d x}{x}

log|log v-2|=log|x|+log |k|

log|log v-2|=log |kx|

log v-2=kx

log v=kx+2

\log \frac{y}{x}=kx+2

\frac{y}{x}=e^{k x+2}

y=xekx+2

Question 36

\left(1-x^{2}\right) \frac{d y}{d x}+x y=a x
Sol :
दोनो तरफ (1-x2) से भाग देने पर

\frac{d y}{d x}+\frac{x y}{1-x^{2}}=\frac{a x}{1-x^{2}}

P=\frac{x}{1-x^{2}},Q=\frac{a x}{1-x^{2}}

Integrating factor=e^{\int P dx}

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

=e^{-\frac{1}{2}} \log\left|1-x^{2}\right|

=e^{\log \left(1-x^{2}\right)^{-\frac{1}{2}}}

=\frac{1}{\left(1-x^{2}\right)^{1 / 2}}=\frac{1}{\sqrt{1-x^{2}}}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

y \cdot \frac{1}{\sqrt{1-x^{2}}}=\int \frac{a x}{1-x^{2}} \times \frac{1}{\sqrt{1-x^{2}}} d x+C

\frac{y}{\sqrt{1-x^{2}}}=a\left.\int \frac{x}{\left(1-x^{2}\right)^{3 / 2}} d x+C\right.

Put 1-x2=t
Differentiating w.r.t x

-2 x=\frac{dt}{dx}

x d x=-\frac{dt}{2}

\frac{y}{\sqrt{1-x^{2}}}=a \int \frac{1}{t^{\frac{3}{2}}}\left(-\frac{dt}{2}\right)+C

\frac{y}{\sqrt{1-x^{2}}}=-\frac{a}{2} \int t^{-\frac{3}{2}} d t+c

\frac{y}{\sqrt{1-x^{2}}}=-\frac{a}{2} \times \frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}+C

\frac{y}{\sqrt{1-x^{2}}}=\frac{a}{\left(1-x^{2}\right)^{\frac{1}{2}}}+C

y=a+c \sqrt{1-x^{2}}

Question 37

\frac{d y}{d x}+y \cot x=2 x+x^{2} \cot x , given that y(0)=0

Sol :

P=cot x , Q=2x+ x2cot x

Integrating factor=e^{\int P dx}

=e^{\int \cot x d x}=e^{\log |\sin x | }
=sin x

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

y \sin x=\int\left(2 x+x^{2} \cot x\right) \cdot \sin x d x+C

y \cdot \sin x=\int 2 x \sin x d x+\int x^{2} \cos x d x+C

y \cdot \sin x=\int 2 x \sin x d x+x^{2} \int \cos x d x-\int \left\{\frac{d\left(x^{2}\right)}{4}\cdot \int \cos xdx \right\}dx+C

y.sin x=∫2x sinx dx+x2sin x-∫2xsinx dx+C

y.sin x=x2sin x+C

At, y(0)=0 या y=0, जब x=0

0 sin 0=02sin 0+C
C=0

∴अवकल समीकरण का हल
ysin x=x2sin x
y=x2

Question 38

ydx+(x-y2)dy=0 , y>0
Sol :
ydx=-(x-y2)dy

\frac{d x}{d y}=-\frac{x+y^{2}}{y}

\frac{d x}{d y}=-\frac{x}{y}+y

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

\frac{d x}{dy}+\left(\frac{1}{y}\right) \cdot x=y

P=\frac{1}{y} , Q=y

Integrating factor=e^{\int P dx}

=e^{\int \frac{1}{y} dy}=e^{\log y}
=y

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

x.y=∫y.ydy+C1
xy=∫y2dy+C1

x y=\frac{y^{3}}{3}+C_1

x y=\frac{y^{3}+3 C_{1}}{3}

3xy=y3+3C1

3xy=y3+C , जहाँ C=3C1

Question 39

\left(x+3 y^{2}\right) \frac{d y}{d x}=y(y>0)
Sol :
\frac{d y}{d x}=\frac{y}{x+3 y^{2}}

\frac{d x}{d y}=\frac{x+3 y^{2}}{y}

\frac{d x}{d y}=\frac{x}{y}+3 y

\frac{d x}{d y}-\frac{x}{y}=3 y

P=-\frac{1}{y} , Q=3y

Integrating factor=e^{\int P dx}

=e^{-\int \frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C_1

x \cdot \frac{1}{y}=\int 3 y \cdot \frac{1}{y} d y+C

\frac{x}{y}=3 y+C

x=3y2+Cy


Question 40

ydy-(x+2y2)dy=0
Sol :
ydx=(x+2y2)dy

\frac{d x}{d y}=\frac{x+2 y^{2}}{y}

\frac{d x}{d y}=\frac{x}{y}+2 y

\frac{d x}{d y}-\frac{x}{y}=2 y

P=-\frac{1}{y} , Q=2y


Question 41

(x+y+1) \frac{d y}{d x}=1
Sol :
\frac{d y}{dx}=\frac{1}{x+y+1}

\frac{dx}{d y}=x+y+1

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

P=-1 , Q=y+1

Integrating factor=e^{\int P dx}
=e^{-\int 1d y}=e^{-y}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

x \cdot e^{-y}=\int(y+1) e^{-y} d y+C

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

x \cdot e^{-y}=-(y+1) e^{-y}+\int e^{-y} d y+C

xe-y=-(y+1)e-y-e-y+C

xe-y=(-y-1-1)e-y+C

x=\frac{(-y-2) e^{-y}+c}{e^{-y}}

x=-(y+2)+Cey

Question 42

\left(x-y^{3}\right) \frac{d y}{d x}+y=0
Sol :
\frac{d y}{dx}=\frac{-y}{x-y^{3}}

\frac{dx}{d y}=\frac{-\left(y^{3}-x\right)}{-y}

\frac{d y}{d x}=y^{2}-\frac{x}{y}

\frac{d y}{d x}+\frac{x}{y}=y^{2}

p=\frac{1}{y},Q=y^{2}

Integrating factor=e^{\int P dx}

=e^{\int \frac{1}{4} d y}=e^{\log y}=y


Question 43

\frac{d y}{d x}+x \sin 2 y=x^{3} \cos ^{2} y
Sol :
\frac{d y}{d x}+2 x \sin y \cos y=x^{3} \cos ^{2} y

Cosy से भाग देने पर

\sec ^{2} y \frac{d y}{d x}+2 x \tan y=x^{3}

Put tan y=v
Differentiating w.r.t. x
\sec ^{2} y \cdot \frac{d y}{d x}=\frac{d v}{d x}

\frac{d v}{d x}+2 x \cdot v=x^{3}

Integrating factor=e^{\int P dx}

=e^{2 \int x dx}=e^{x^{2}}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

v.e^{x^{2}}=\int x^{3} \cdot e^{x^{2}} d x+C

Put x2=t
Differentiating w.r.t x
2 x=\frac{dt}{dx}
x d x=\frac{d t}{2}

v \cdot e^{x^{2}}=\int t \cdot e^{t} \frac{d t}{2}+C

v \cdot e^{x^{2}}=\frac{1}{2} \int t \times e^{t} d t+C

v e^{x^{2}}=\frac{1}{2}\left[t \int e^{t} d t-\int\left\{\frac{d(t)}{d t} \cdot \int e^{t} dt\right\} d t\right]+C

ve^{x^{2}}=\frac{1}{2}\left[t e^{t}-e^{t}\right]+C

ve^{x^{2}}=\frac{1}{2}(t-1) e^{t}+C

ve^{x^{2}}=\frac{1}{2}\left(x^{2}-1\right) e^{x^{2}}+C

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

Question 44

x \frac{d y}{d x}+y=y^{2} \log x
Sol :
Divide by xyon both sides

\frac{x}{x y^{2}} \frac{d y}{dx}+\frac{y}{x y^{2}}=\frac{y^{2} \log x}{xy^{2}}

\frac{1}{y^{2}} \frac{d y}{dx}+ \frac{1}{x y}=\frac{\log x}{x}

Put \frac{1}{y}=v
Differentiating w.r.t x
-\frac{1}{y^{2}} \frac{d y}{dx}=\frac{d v}{d x}
\frac{1}{y^{2}} \frac{dy}{dx}=-\frac{d v}{dx}


-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x}

\frac{dv}{dx}-\frac{v}{x}=-\frac{\log _{x}}{x}

p=-\frac{1}{x},Q=-\frac{\log x}{x}

Integrating factor=e^{\int P dx}

=e^{-\int \frac{1}{x} dx}=e^{-\log x}=\frac{1}{x}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

v\cdot \frac{1}{x}=-\int \frac{\log x}{x} \times \frac{1}{x} d x+C

v \frac{1}{x}=-\int \operatorname{log} x \cdot \frac{1}{x^{2}} d x+C

\frac{v}{x}=\int \log x \cdot\left(-\frac{1}{x^{2}}\right) d x+c

\frac{v}{x}=\log x\int\left(\frac{-1}{x^{2}}\right)dx-\int \left\{\frac{d(\log x)}{dx} \int \left(-\frac{1}{x^2}\right)dx\right\}dx

\frac{v}{x}=\log x \times \frac{1}{2}-\int \frac{1}{x} \times \frac{1}{x} d x+c

\frac{v}{x}=\frac{\log x}{x}+\frac{1}{x}+C

\frac{1}{y x}=\frac{\log x+1}{x}+C

Question 45

\frac{d y}{d x}=x^{3} y^{3}-x y
Sol :
\frac{dy}{d x}+x y=x^{3} y^{3}

yसे भाग देने पर

\frac{1}{y^3} \frac{d y}{d x}+\frac{xy}{y^{3}}=\frac{x^{3} y^{3}}{y^{3}}

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

Put \frac{1}{y^{2}}=v
y-2 =v
Differentiating w.r.t x
-2 y^{-3} \frac{d y}{d x}=\frac{d v}{d x}
\frac{1}{y^{3}} \frac{d y}{d x}=-\frac{1}{2} \frac{d v}{dx}

-\frac{1}{2} \frac{d v}{dx}+v x=x^{3}

-2 से गुणा करने पर

\frac{d v}{dx}-2 v x=-2 x^{3}

P=-2x, Q=-2x3

Integrating factor=e^{\int P dx}

=e^{-\int 2 xdx}=e^{-x^{2}}

दिए गए अवकल समीकरण का हल

y.Integral factor==\int Q \cdot(\text{Integral factor}) d x+C

v \cdot e^{-x^{2}}=\int\left(-2 x^{3}\right) \cdot e^{-x^{2}} d x+C

Put -x2=t
Differentiating w.r.t x
-2x dx=dt

\frac{v}{e^{x^{2}}}=-\int te^{t} d t+C

\frac{v}{e^{x^{2}}}=-\left[t \int e^{t} d t-\int\left\{\frac{d(t)}{dt} \int e^{t} d t\right\} dt\right]+C

\frac{v}{e^{x^{2}}}=-\left[t e^{t}-e^{t}\right]+C

\frac{u}{e^{x^{2}}}=-e^{x}(t-1)+C

\frac{v}{e^{x^{2}}}=-e^{-x^{2}}\left(-x^{2}-1\right)+C

v=-1\left(-x^{2}-1\right)+Ce^{x^{2}}

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

1=\left(x^{2}+1\right) y^{2}+Cy^{2} e^{x^{2}}

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