KC Sinha: Exercise 23.7- Mathematics Solution Class 12 Chapter 23 अवकल समीकरण

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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.e^2x=e^x+C

Question 8

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

Sol :

P=2 , Q=6e^x

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

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.e^2x=2.e^3x+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$=e^{\int Pdx}=e^{\int 2 dx}$=e^2x

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=e^x

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

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

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$=e^{\int Pdx}=e^{\int 3 d x}$=e^3x

दिए गए अवकल समीकरण का हल:

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

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

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

ye^3x=e^x+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

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

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=e^x(sin x-cos x)-I

2I=e^x(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=e^x

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

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

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

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

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

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

ye^x=e^x cos 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=e^x sec 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=e^x+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)+Ce^{sin x}

y=Ce^{sin 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.e^{sin x}$=\int e^{t} d t+c$

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

y.e^{sin x}=e^t.t-e^t+C

y.e^{sin x}=e^t(t-1)+C

y.e^{sin x}=e^{sin 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=e^x

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=xe^x-e^x+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=xe^kx+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+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)+Ce^y

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 :

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

Cos² y 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 xy² on 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 :

$\frac{dy}{d x}+x y=x^{3} y^{3}$

y³ से 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

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

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

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

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

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

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

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

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