KC Sinha Solution Class 12 Chapter 23 अवकल समीकरण (Differential Equations) Exercise 23.6

 Exercise 23.6

निम्नलिखित अवकल समीकरणो को हल करे।

[Solve the following differential equations]

Question 1

x(x-y)dy+y²dx=0

Sol :

x(x-y)dy=-y²dx

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

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

f(x,y)$=\frac{dy}{dx}=\frac{y^2}{xy-x^2}$...(i)

f(kx,ky)$=\frac{(ky)^2}{kx.ky}-(kx)^2$

$=\frac{k^2y^2}{k^2xy-k^2x^2}$

$=\frac{k^2y^2}{k^2(xy-x^2)}$

=k⁰f(x,y)

∴यह एक समघातीय अवकल समीकरण है।

y=vx

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii) 

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{y^2}{xy-x^2}$

$\frac{dv}{dx}.x+v=\frac{(vx)^2}{x.vx-x^2}$

$\frac{dv}{dx}.x+v=\frac{v^2x^2}{x^2(v-1)}$

$\frac{dv}{dx}.x=\frac{v^2}{v-1}-v$

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

$\frac{v-1}{v}dv=\frac{dx}{x}$

$\left(1-\frac{1}{v}\right)dv=\frac{dx}{x}$

Integrating both sides

$\int \left(1-\frac{1}{v}\right)dv=\int \frac{dx}{x}$

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

$\frac{y}{x}=\log \left|\frac{y}{x}\right|+\log |x|+\log k$

$\frac{y}{x}=\log \left|\frac{y}{x}.x.k\right|$

$e^{\frac{y}{x}}=ky$

$y=\frac{1}{k}e^{\frac{y}{x}}$

$y=ce^{\frac{y}{x}}$ जहाँ $c=\frac{1}{k}$

Question 2

x²dy+y(x+y)dx=0

Sol :

x²dy+y(x+y)dx=0

x²dy=-y(x+y)dx

$\dfrac{dy}{dx}=\frac{-y(x+y)}{x^2}$

$\frac{dy}{dx}=-\left(\frac{xy+y^2}{x^2}\right)$...(i)

∴यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=-\left(\frac{xy+y^2}{x^2}\right)$

$\frac{dv}{dx}.x+v=-\left(\frac{x.vx+(vx)^2}{x^2}\right)$

$\frac{dv}{dx}.x+v=-\left(\frac{x^2(v+v^2)}{x^2}\right)$

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

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

$\frac{dv}{dx}.x=-(v^2+2v$

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

$\frac{dv}{v(v+2)}=-\frac{dx}{x}$

Integrating both sides

$\int \frac{dv}{v(v+2)}=-\int \frac{dx}{x}$

$\frac{1}{2}\int \left[\frac{1}{v}-\frac{1}{v+2}\right]dv=-\int \frac{dx}{x}$

$\frac{1}{2}\left[\log|v|-\log|v+2|\right]=-\log|x|+\log k$

$\log \frac{v}{v+2}$=-2log|x|+2logk

$\log \dfrac{\frac{y}{x}}{\frac{y}{x}+2}+\log x^2 =\log k^2$

$\log \dfrac{\frac{y}{x}}{\frac{y+2x}{x}}+\log x^2 =\log k^2$

$\log \frac{x^2 y}{y+2x}=\log k^2$

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

x²y=k²(y+2x)

x²y=c(y+2x), जहाँ c=k²

Question 3

$x\frac{dy}{dx}=x+y$

Sol :

$\frac{dy}{dx}=\frac{x+y}{x}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$...(ii)

समीकरण (i) तथा (ii) से

$\frac{dv}{dx}x+v=\frac{x+y}{x}$

$\frac{dv}{dx}.x+v=\frac{x+vx}{x}$

$\frac{dv}{dx}.x+v==\frac{x(1+v)}{x}$

$\frac{dv}{dx}.x+v=1+v-v$

$dv=\frac{dx}{x}$

Integrating both sides

$\int dv=\int \frac{dx}{x}$

v=log x+c

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

y=xlog|x|+cx

Question 4

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Sol :

$\frac{dy}{dx}=\frac{x+y}{x-y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{x+y}{x-y}$

$\frac{dv}{dx}.x+v=\frac{x+vx}{x-vx}$

$\frac{dv}{dx}.x+v=\frac{x(1+v)}{x(1-v)}$

$\frac{dv}{dx}.x=\frac{1+v}{1-v}-v$

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

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

Integrating both sides

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

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

$\int \frac{1}{1+v^2}dv-\frac{1}{2}\int \frac{2v}{1+v^2}dv=\int \frac{dv}{x}$

$\tan^{-1}v-\frac{1}{2}\log |1+v^2|=\log x+c$

$\tan^{-1}v=\frac{1}{2}\log |1+v^2|+\log x+c$

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

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

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

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

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

Question 5

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

Sol :

$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{x^2+y^2}{2xy}$

$\frac{dv}{dx}.x+v=\frac{x^2+(vx)^2}{2x.vx}$

$\frac{dv}{dx}.x+v=\frac{x^2(1+v^2)}{2x^2v}$

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

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

$\frac{dv}{dx}.x=\frac{1-v^2}{2v}$

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

Integrating both sides

$-\int \frac{-2v}{1-v^2}dv=\int \frac{dx}{x}$

-log|1-v²|=log |x|+log k

$-\log \left|1-\frac{y^2}{x^2}\right|=\log |x|+\log k$

$-\log \left|\frac{x^2-y^2}{x^2}\right|=\log |x|+\log k$

-log|x²-y²|+log x²=log x+log k

2log x-log x=log(x²-y²)+log k

log x=log k(x²-y²)

x=k(x²-y²)

Question 6

ye^x/ydx=(xe^x/y+y)dy

Sol :

$f(x,y)=\frac{dx}{dy}=\frac{xe^{x/y}+y}{ye^{x/y}}$..(i)

$f(kx,ky)=\frac{kxe^{\frac{kx}{ky}}+ky}{kye^{\frac{kx}{ky}}}$

$=\frac{k(xe^{\frac{x}{y}}+y)}{kye^{x/y}}$

=k⁰.f(x,y)

यह एक समघातीय अवकल समीकरण है।

x=vy ⇒ $v=\frac{x}{y}$

Differentiating w.r.t x

$\frac{dx}{dy}=\frac{dv}{dy}.y+v$..(ii)

समीकरण (i) तथा (ii) से,

 $\frac{dv}{dy}.y+v=\frac{x.e^{\frac{x}{y}}+y}{ye^{\frac{x}{y}}}$

$\frac{dv}{dy}.y+v=\frac{vye^{\frac{vy}{y}}+y}{ye^{\frac{vy}{y}}}$

$\frac{dv}{dy}.y+v=\frac{y(ve^v+1)}{ye^v}$

$\frac{dv}{dy}.y=\frac{ve^v+1}{e^v}-v$

$\frac{dv}{dy}.y=\frac{ve^v+1-ve^v}{e^v}$

$e^vdv=\frac{dy}{y}$

Integrating both sides

$\int e^v dv=\int \frac{dy}{y}$

e^v=log|y|+c

$e^{\frac{x}{y}}=\log y+c$

Question 7

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

Sol :

$\frac{dy}{dx}=\frac{x^2-y^2}{xy}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से

$\frac{dv}{dx}.x+v=\frac{x^2-y^2}{xy}$

$\frac{dv}{dx}.x+v=\frac{x^2-(vx)^2}{x.vx}$

$\frac{dv}{dx}.x+v=\frac{x^2(1-v^2)}{x^2 v}$

$\frac{dv}{dx}.x=\frac{1-v^2}{v}-v$

$\frac{dv}{dx}.x=\frac{1-v^2-v^2}{v}-v$

$\frac{dv}{dx}.x=\frac{1-2v^2}{v}$

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

Integrating both sides

$\frac{1}{-4}\int \frac{-4v}{1-2v^2}dv=\int \frac{dx}{x}$

$-\frac{1}{4}\log |1-2v^2|=\log|x|+\log k$

$\log |1-2\frac{y^2}{x^2}|=-4\log |x|-4\log k$

$\log \left|\frac{x^2-2y^2}{x^2}\right|=-\log |x^4|+\log k^{-4}$

$\log \left|\frac{x^2-2y^2}{x^2}\times x^4\right|=\log k^{-4}$

log|x²(x²-2y²)|=log k^-4

x²(x²-2y²)=k^-4

x²(x²-2y²)=c, जहाँ c=k^-4

Question 8

$\frac{dy}{dx}=\frac{x-y}{x+y}$

Sol :

$\frac{dy}{dx}=\frac{x-y}{x+y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒$v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}x+v=\frac{x-y}{x+y}$

$\frac{dv}{dx}x+v=\frac{x-vx}{x+vx}$

$\frac{dv}{dx}x+v=\frac{x(1-v)}{x(1+v)}$

$\frac{dv}{dx}.x=\frac{1-v}{1+v}-v$

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

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

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

Integrating both sides

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

$-\frac{1}{2}\log |1-2v-v^2|=\log |x|+\log k$

$\log \left|1-\frac{2y}{x}-\frac{y^2}{x^2}\right|=-2\log |x|-2\log k$

$\log \left|\frac{x^2-2xy-y^2}{x^2}\right|=-\log |x^2|+\log k^{-2}$

$\log \left|\frac{x^2-2xy-y^2}{x^2}\times x^2\right|=\log \frac{1}{k^2}$

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

$x^2-2xy-y^2=c$ , जहाँ $c=\frac{1}{k^2}$

Question 9

$2xy\frac{dy}{dx}=x^2+3y^2$

Sol :

$\frac{dy}{dx}=\frac{x^2+3y^2}{2xy}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}x+v=\frac{x^2+3y^2}{2xy}$

$\frac{dv}{dx}x+v=\frac{x^2+3(vx)^2}{2x.vx}$

$\frac{dv}{dx}x+v=\frac{x^2(1+3v^2)}{2x^2v}$

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

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

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

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

Integrating both sides

$\int \frac{2v}{1+v^2}dv=\int \frac{dx}{x}$

log|1+v²|=log|x|+log c

$\log \left|1+\frac{y^2}{x^2}\right|=\log x+\log c$

$\log \left|\frac{x^2+y^2}{x^2}\right|=\log cx$

$\frac{x^2+y^2}{x^2}=cx$

x²+y²=cx³

Question 10

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

Sol :

$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}.x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}x+v=\frac{2xy}{x^2-y^2}$

$\frac{dv}{dx}x+v=\frac{2x.vx}{x^2-(vx)^2}$

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

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

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

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

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

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

Integrating both sides

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

$\int \left[\frac{1}{v}-\frac{2v}{1+v^2}\right]=\int \frac{dx}{x}$

log|v|-log|1+v²|=log|x|+log k

$\log \frac{v}{1+v^2}=\log x+\log k$

$\log \dfrac{\frac{y}{x}}{1+\frac{y^2}{x^2}}=\log kx$

$\dfrac{\frac{y}{x}}{\frac{x^2+y^2}{x^2}}=kx$

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

y=k(x²+y²)

Question 11

$\frac{dy}{dx}+\frac{x-2y}{2x-y}=0$

Sol :

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

$\frac{dy}{dx}=\frac{2y-x}{2x-y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{2y-x}{2x-y}$

$\frac{dv}{dx}x+v=\frac{2vx-x}{2x-vx}$

$\frac{dv}{dx}.x+v=\frac{x(2v-1)}{x(2-v)}$

$\frac{dv}{dx}.x=\frac{2v-1}{2-v}-v$

$\frac{dv}{dx}.x=\frac{v^2-1}{2-v}$

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

Integrating both sides

$\int \frac{2-v}{v^2-1}dv=\int \frac{dx}{x}$

$2\int \frac{dv}{v^2-1^2}-\frac{1}{2}\int \frac{v}{v^2-1}dv=\int \frac{dx}{x}$

$2\times \frac{1}{2\times 1} \log \left|\frac{v-1}{v+1}\right|-\frac{1}{2}\log \left|v^2-1\right|=\log |x|+\log k$

$\log \left|\frac{v-1}{v+1}\right|-\log |\sqrt{v^2-1}|$=log kx

$\log \left|\frac{v-1}{(v+1)\sqrt{(v-1)(v+1)}}\right|=\log kx$

$\log \left|\frac{\sqrt{v-1}}{(v+1)\sqrt{v+1}}\right|=\log kx$

$\dfrac{\sqrt{\frac{y}{x}-1}}{\left(\frac{y}{x}+1\right)\sqrt{\frac{y}{x}+1}}=kx$

दोनो तरफ वर्ग करने पर,

$\dfrac{\frac{y-x}{x}}{\frac{(x+y)^3}{x^3}}=k^2x^2$

$\frac{x^2(y-x)}{(x+y)^3}=k^2 x^2$

y-x=k²(x+y)³

y-x=c(x+y)³, जहाँ c=k²

Question 12

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

Sol :

$\frac{dy}{dx}=\frac{x^2+xy+y^2}{x^2}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{x^2+xy+y^2}{x^2}$

$\frac{dv}{dx}.x+v=\frac{x^2+x.vx+(vx)^2}{x^2}$

$\frac{dv}{dx}.x+v=\frac{x^2(1+v+v^2)}{x^2}$

$\frac{dv}{dx}.x+v=1+v+v^2$

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

Integrating both sides

$\int \frac{dv}{1+v^2}=\int \frac{dx}{x}$

tan^-1 v=log|x|+c

$\tan \left(\frac{1}{x}\right)=\log |x|+C$

Question 13

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

Sol :

$\frac{dy}{dx}=\frac{y-\sqrt{x^2y^2}}{x}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{dx}.x+v=\frac{y-\sqrt{x^2+y^2}}{x}$

$\frac{dv}{dx}.x+v=\frac{vx-\sqrt{x^2+(vx)^2}}{x}$

$\frac{dv}{dx}.x+v=\frac{vx-x\sqrt{1+v^2}}{x}$

$\frac{dv}{dx}.x+v=\frac{x(v-\sqrt{1+v^2})}{x}$

$\frac{dv}{dx}.x+v=v-\sqrt{1+v^2}$

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

Integrating both sides

$\int \frac{dv}{\sqrt{1^2+v^2}}=-\int \frac{dx}{x}$

$\log (v+\sqrt{1^2+v^2})=-\log|x|+\log k$

$\log \left(\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right)+\log x=\log k$

$\log \left(\frac{y}{x}+\sqrt{\frac{x^2+y^2}{x^2}}\right)+\log x=\log k$

$\log \left(\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x}\right)+\log x=\log k$

$\log \left(\frac{y+\sqrt{x^2+y^2}}{x}\right)+\log x=\log k$

$\log \left(\frac{y+\sqrt{x^2+y^2}}{x}\times x\right)=\log k$

$y+\sqrt{x^2+y^2}=k$

Question 14

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

Sol :

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

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

$\frac{dy}{dx}=\frac{y^2}{xy-x^2}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

$\frac{d v}{d x} \cdot x+v=\frac{(v x)^{2}}{x \cdot v x-x^{2}}$

$\frac{d v}{d x} \cdot x+v=\frac{v^{2} x^{2}}{x^{2}(v-1)}$

$\frac{d v}{d x} \cdot x=\frac{v^{2}}{v-1}-v$

$\frac{d v}{d x} x=\frac{v^{2}-v^{2}+v}{u-1}$

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

$\left(1-\frac{1}{v}\right) d v=\frac{d x}{x}$

Integrating both sides

$\int\left(1-\frac{1}{v}\right) d v=\int \frac{d x}{x}$

v-log v=log x+c

$\frac{y}{x}-\log \frac{y}{x}=\log x+c$

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

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

y=x[log y+C]

Question 15

(x²+)dx+3xydy=0

3xydy=-(x²+y²)dx

$\frac{d y}{d x}=-\frac{\left(x^{2}+y^{2}\right)}{3 x y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=-\frac{\left(x^{2}+y^{2}\right)}{3 x y}$

$\frac{d v}{d x} \cdot x+v=-\frac{\left(x^{2}+v^{2} x^{2}\right)}{3 x \cdot v x}$

$\frac{d v}{d x} \cdot x+v=\frac{-x^{2}\left(1+v^{2}\right)}{3 vx^{2}}$

$\frac{d v}{d x} \cdot x=\frac{-1-v^{2}}{3 v}-v$

$\frac{d v}{d x} \cdot x=\frac{-1-v^{2}-3 v^{2}}{3 v}$

$\frac{d v}{d x} \cdot x=\frac{-1-4 v^{2}}{3 v}$

$\frac{d v}{d x} \cdot x=-\frac{\left(1+4 v^{2}\right)}{3 v}$

$\frac{3 v}{1+4 v^{2}} d v=-\frac{d x}{x}$

Integrating both sides

$3\int {\frac{v}{1+4 v^{2}}} d v=-\int \frac{dx}{x}$

$\frac{3}{8} \int \frac{8 v}{1+4 v^{2}} d v=-\int \frac{dx}{x}$

$\frac{3}{8} \operatorname{log}\left(1+4 v^{2}\right)$=-log|x|+log k

$\frac{3}{8} \log\left|1+4 \frac{y^{2}}{x^{2}}\right|$=-log|x|+log k

$\log \left| \frac{x^{2}+4 y^{2}}{x^{2}}\right|=-\frac{8}{3} \log |x|+\frac{8}{3} \log k$

$\log \left|\frac{x^{2}+4 y^{2}}{x^{2}}\right|+\log x^{\frac{8}{3}}=\log {k^{\frac{8}{3}}}$

$\log \left(\frac{x^{2}+4 y^{2}}{x^2} \times x^{\frac{8}{3}}\right)=\log k^{\frac{8}{3}}$

$x^{2 / 3}\left(x^{2}+4 y^{2}\right)=k^{8 / 3}$

दोनो तरफ घन करने पर,

x²(x²+4y²)³=k⁸

x²(x²+4y²)³=c ,जहाँ c=k⁸

Question 16

x(x-y)dy+y²dx=0

Sol :

(x²-xy)dy=-y²dx

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

$\frac{d y}{d x}=\frac{y^{2}}{x y-x^{2}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

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

$\frac{d v}{d x} \cdot x+v=\frac{v^{2} x^{2}}{x^{2}(v-1)}$

$\frac{d v}{x} \cdot x=\frac{v^{2}}{v-1}-v$

$\frac{d v}{d x} \cdot x=\frac{v^{2}-v^{2}+v}{v-1}$

$\frac{v-1}{v} d v=\frac{dx}{x}$

$\left(\frac{v}{v}-\frac{1}{v}\right) d v-\frac{dx}{x}$

$\left(1-\frac{1}{v}\right) d v=\frac{dx}{x}$

Integrating both sides

$\int\left(1-\frac{1}{v}\right) d v=\int \frac{d x}{x}$

v-log v=log x+ log k

v=log v+ log x+ log k

$\frac{y}{x}=\log \frac{y}{x}+\log {x}+\log k$

$\frac{y}{x}=\log \frac{y}{x} \cdot x \cdot k$

$\frac{y}{x}=\log ky$

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

$\frac{1}{k} e^{y / x}=y$

$c e^{y /x}=y$, जहाँ $c=\frac{1}{k}$

Question 17

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

Sol :

$\frac{d y}{d x}=\frac{x+2 y}{x-y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{x+2 y}{x-y}$

$\frac{d v}{d u} \cdot x+v=\frac{x+2 v x}{x-v x}$

$\frac{d v}{d x} x+v=\frac{x(1+2 v)}{x(1-v)}$

$\frac{d v}{d x} x=\frac{1+2 v}{1-v}-v$

$\frac{d v}{d x} \cdot x=\frac{1+2 v-v+v^{2}}{1-v}$

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

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

Integrating both side

$\int \frac{1-v}{v^{2}+v+1} d v=\int \frac{d x}{x}$

$1-v=\frac{A \cdot d\left(u^{2}+u+1\right)}{dv}+B$

1-v=A(2v+1)+B

1-v=2AV+A+B

By equating coefficient of v and constant 

2A=-1

$A=-\frac{1}{2}$

A+B=1

$-\frac{1}{2}+B=1 \Rightarrow B=\frac{3}{2}$

$\int \frac{A(2 v+1)+B}{v^{2}+v+1} d v=\int \frac{dx}{x}$

$A\int \frac{2 v+1}{v^{2}+v+1} d v+B \int \frac{1}{v^{2}+v+1} d v=\int \frac{d x}{x}$

v²+v+1$=v^{2}+2 \cdot v \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}+1-\left(\frac{1}{2}\right)^{2}$

$=\left(v+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^{2}$

$-\frac{1}{2} \int \frac{2 v+1}{u^{2}+v+1} d v+\frac{3}{2} \int \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(v+\frac{1}{2}\right)^{2}} d v=\int \frac{d x}{x}$

$-\frac{1}{2} \log\left|v^{2}+v+1\right|+\frac{3}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1} \frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+c=\log |x|$

$\sqrt{3}\tan ^{-1} \frac{\frac{2 v+1}{2}}{\frac{\sqrt{3}}{2}}+c=\frac{1}{2} \log \left|v^{2}+v+1\right|+\log |x|$

$\sqrt{3} \tan ^{-1} \dfrac{\frac{2y}{x}+1}{\sqrt{3}}+C=\frac{1}{2}\left[\log \left|\frac{y^{2}}{x^{2}}+\frac{y}{2}+1\right|+2\log |x| \right]$

$2 \sqrt{3}\tan^{-1} \frac{\frac{2 y+x}{x}}{\sqrt{3}}+c=\log \left| \frac{y^{2}+x y+x^{2}}{x^{2}}\right|+\log|x^2|$

$2 \sqrt{3} \tan ^{-1} \frac{x+2 y}{\sqrt{3} x}+c=\log \left| \frac{y^{2}+x y+x^{2}}{x^{2}} \times x^{2}\right|$

$2 \sqrt{3} \tan ^{-1} \frac{x+2 y}{\sqrt{3} x}+c=\log \left|x^{2}+x y+y^{2}\right|$

Question 18

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

Sol :

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

$f\left(k_{x}, k_{y}\right)=\frac{ky\left(\log k y-\log kx+1\right)}{k x}$

$=\frac{y\left(\operatorname{log} \frac{k y}{k x}+1\right)}{x}$

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

=k⁰ f(x,y)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

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

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

$\frac{d v}{dx} \cdot x+v=v \log v+v$

$\frac{d v}{v \log v}=\frac{d x}{x}$

Integrating both sides

$\int \frac{d v}{v \log v}=\int \frac{d x}{x}$

log|log v|=log |x|+log C

$\log \left|\log \frac{y}{x}\right|=\log | c x|$

$\log \frac{y}{x}=C x$

$\frac{y}{x}=e^{Cx}$

y=xe^Cx

Question 19

$(x-y) \frac{d y}{d x}=x+3 y$

Sol :

$\frac{d y}{d x}=\frac{x+3 y}{x-y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{dx} \cdot x+v=\frac{x+3 y}{x-y}$

$\frac{d v}{d x} \cdot x+v=\frac{x+3 v x}{x-v x}$

$\frac{d v}{d x} x+v=\frac{x(1+3 v)}{x(1-v)}$

$\frac{d v}{d x} \cdot x=\frac{1+3 v}{1-v}-v$

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

$\frac{d v}{d x} \cdot x=\frac{v^{2}+2v+1}{1-v}$

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

Integrating both sides

$\int \frac{1-v}{v^{2}+2 v+1} d v=\int \frac{d x}{x}$

$1-v=\frac{A \cdot d\left(v^{2}+2 v+1\right)}{d v}+B$

1-v=A(2v+2)+B

Question 20

(x³+3xy²)dx+(y³+3x²y)dy=0

Sol :

(y³+3x²y)dy=-(x³+3xy²)dx

$\frac{d y}{d x}=-\frac{\left(x^{3}+3 x y^{2}\right)}{y^{3}+3 x^{2} y}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=-\frac{\left(x^{3}+3 x y^{2}\right)}{y^{3}+3 x^{2} y}$

$\frac{d v}{d x} \cdot x+v=\frac{-\left(x^{3}+3 x v^{2} x^{2}\right)}{v^{3} x^{3}+3 x^{2} \cdot v x}$

$\frac{dv}{d x} \cdot x+v=-\frac{x^{3}\left(1+3 v^{2}\right)}{x^{3}\left(v^{3}+3 v\right)}$

$\frac{d v}{d x} \cdot x=-\frac{\left(1+3 v^{2}\right)}{v^{3}+3 v}-v$

$\frac{d v}{dx} \cdot x=\frac{-1-3 v^{2}-v^{4}-3 v^{2}}{u^{3}+3 v}$

$\frac{d v}{dx} \cdot x=\frac{-v^{4}-6 v^{2}-1}{v^{3}+3 v}$

$\frac{v^{3}+3 v}{v^{4}+6 v^{2}+1} d v=-\frac{d x}{x}$

Integrating both sides

$\int \frac{v^{3}+3 v}{v^{4}+6 v^{2}+1} d v=-\int \frac{d x}{x}$

$\frac{1}{4} \int \frac{4 v^{3}+12 v}{v^{4}+6 v^{2}+1} d v=-\int \frac{dx}{x}$

$\frac{1}{4} \log \left|v^{4}+6 v^{2}+1\right|=-\log|x|+\log k$

log |v⁴+6v²+1|=-4log |x|+4log k

$\log \left|\frac{y^{4}}{x^{4}}+\frac{6 y^{2}}{x^{2}}+1\right|+\operatorname{log}\left|x^{4}\right|=\log k^{4}$

$\log \left|\frac{y^{4}+6 x^{2} y^{2}+x^{4}}{x^{4}}\right|+\log |x^{4}| =\log k^{4}$

$\log \left|\frac{y^{4}+6 x^{2} y^{2}+x^{4}}{x^{4}} \times x^{4}\right| =\log k^{4}$

x⁴+6x²y²+y⁴=k⁴

x⁴+6x²y²+y⁴=C, जहाँ C=k⁴

Question 21

$(x-\sqrt{x y}) d y=y d x$

Sol :

$\frac{d y}{d x}=\frac{y}{x-\sqrt{x y}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{dx} \cdot x+v=\frac{y}{x-\sqrt{x y}}$

$\frac{d v}{d x} \cdot x+v=\frac{v x}{x-\sqrt{x \cdot v k}}$

$\frac{dv}{d x} \cdot x+v=\frac{v x}{x(1-\sqrt{v})}$

$\frac{d v}{dx} \cdot x=\frac{v}{1-\sqrt{v}}-v$

$\frac{d v}{d x} \cdot x=\frac{v-v+v \sqrt{v}}{1-\sqrt{v}}$

$\frac{1-\sqrt{v}}{v \sqrt{v}} d v=\frac{d x}{x}$

$\left(\frac{1}{v \sqrt{v}}-\frac{\sqrt{v}}{v\sqrt{v}}\right) d v=\frac{d x}{x}$

$\left(v^{-\frac{3}{2}}-\frac{1}{v}\right) d v=\frac{dx}{x}$

Integrating both sides

$\int\left(v^{\frac{-3}{2}}-\frac{1}{v}\right) d v=\int \frac{d x}{x}$

$\frac{v^{-\frac{1}{2}}}{\frac{-1}{2}}-\log |v|=\log x+\log k$

$-2 v^{-\frac{1}{2}}-\log v=\log k x$

$-\left[2\left(\frac{y}{x}\right)^{-\frac{1}{2}}-\log \frac{y}{x}\right]=\log k x$

$2 \sqrt{\frac{x}{y}}+\log \frac{y}{x}=-\log k x$

$2 \sqrt{\frac{x}{y}}+\log y-\log x=\log (k x)^{-1}$

$2 \sqrt{\frac{x}{y}}+\log y=\log \frac{1}{kx}+\log x$

$2 \sqrt{\frac{x}{y}}+\log y=\log \frac{1}{k}$

$2 \sqrt{\frac{x}{y}}+\log y=C$ , जहाँ $C=\log \frac{1}{k}$

Question 22

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

Sol :

x से भाग देने पर

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

$\frac{d y}{d x}=\frac{y}{x}-\frac{y^{2}}{x^{2}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{dx} \cdot x+v=\frac{y}{x}-\frac{y^{2}}{x^{2}}$

$\frac{d v}{dx} \cdot v+v=v-v^{2}$

$\frac{d v}{v^{2}}=-\frac{d x}{x}$

Integrating both sides

$\int \frac{d v}{v^{2}}=-\int \frac{d x}{x}$

$-\frac{1}{v}=-\log \cdot|x|+c$

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

$\log |x|-\frac{x}{y}=c$

Question 23

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

Sol :

$\frac{d v}{dx}=\frac{2 \sqrt{y^{2}-x^{2}}+y}{x}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{2 \sqrt{y^{2}-x^{2}}+y}{x}$

$\frac{d v}{d x} \cdot x+v=\frac{2 \sqrt{v^{2} x^{2}-x^{2}}+v x}{x}$

$\frac{d v}{d x} \cdot x+v=\cdot \frac{x\left(2 \sqrt{v^{2}-1}+v\right)}{x}$

$\frac{d v}{d x} \cdot x+v=2 \sqrt{{v^{2}-1}}+v$

$\frac{d v}{\sqrt{v^{2}-1}}=2 \frac{d x}{x}$

Integrating both sides

$\int \frac{d v}{\sqrt{v^{2}-1}}=2 \int \frac{d v}{x}$

$\log|v+\sqrt{v^2-1}|$=2log|x|+log C

$\log \left|\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}-1}\right|=\log \left|x^{2}\right|+\log C$

$\log \left|\frac{y}{x}+\sqrt{\frac{y^{2}-x^{2}}{x^{2}}}\right|=\log \left|c x^{2}\right|$

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

$y+\sqrt{y^{2}-x^{2}}=c x^{3}$

Question 24

y²dx+(x²+xy+y²)dy=0

Sol :

(x²+xy+y²)dy=-y²dx

$\frac{d y}{d x}=\frac{-y^{2}}{x^{2}+x y+y^{2}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{dx} \cdot x+v=\frac{-y^{2}}{x^{2}+x y+y^{2}}$

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

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

$\frac{d v}{d x} x=\frac{-v^{2}}{1+u+v^{2}}-v$

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

$\frac{d v}{d x} \cdot x=\frac{-\left(v^{3}+2 v^{2}+v\right)}{v^{2}+v+1}$

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

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

Integrating both sides

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

$\int\left[\frac{v(v+1)}{v(v+1)^{2}}+\frac{1}{v(v+1)^{2}}\right] d v=-\int \frac{d x}{x}$

$\int\left[\frac{1}{v+1}+\frac{1}{v(v+1)^{2}}\right] d v=-\frac{d x}{x}$

Let, 

$\frac{1}{v(v+1)^{2}}=\frac{A}{v}+\frac{B}{(v+1)}+\frac{C}{(v+1)^{2}}$

$\frac{1}{v(v+1)^{2}}=\frac{A(v+1)^{2}+B v(v+1)+C v}{v(v+1)^{2}}$

1=A(v²+2v+1)+B(v²+v)+Cv

A+B=0, 2A+B+C=0, A=1

B=-1, 

2(1)-1+C=0

C=-1

$\int\left[\frac{1}{v+1}+\frac{1}{v} -\frac{1}{v+1} \frac{-1}{(v+1)^{2}}\right] d v=-\int\frac{d x}{x}$

$\log v+\frac{1}{v+1}=-\log x+c$

$\log \frac{y}{x}+\frac{1}{\frac{y}{x}+1}+\log x=c$

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

Question 25

$\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x$

Sol :

$\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}}$

$\frac{d v}{dx} \cdot x+v=\frac{v x+\sin \frac{v x}{x}-x}{x \operatorname{sin} \frac{v x}{x}}$

$\frac{d v}{d x}x+v=\frac{x(v \sin v-1)}{x \sin v}$

$\frac{d v}{d v} \cdot x+v=\frac{v \sin v}{\sin v}-\frac{1}{\sin v}$

$\frac{d v}{dx} \cdot x+v=v-\frac{1}{\sin v}$

$-\sin v d v=\frac{d x}{x}$

Integrating both sides

$-\int \sin v d v=\int \frac{dx}{x}$

cos v=log |x|+C

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

Question 26

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

Sol :

$\frac{d y}{d x}=\frac{y}{x}+\sin \frac{y}{x}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{dv}{d x} x+v=\frac{y}{x}+\sin\frac{y}{x}$

$\frac{d v}{d x} \cdot x+v=v+\sin v$

$\frac{d v}{\sin v}=\frac{dx}{x}$

Integrating both sides

$\int \operatorname{cosec} v d v=\int \frac{d x}{x}$

$\log \left|1+\frac{v}{2}\right|=\log | x \mid+\log c$

$\log \left|\frac{y}{2 x}\right|=\log | c x \mid$

$\tan \left(\frac{y}{2x}\right)=C x$

Question 27

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

Sol :

$\frac{d y}{d x}=\frac{y}{2}-\cos ^{2}\left(\frac{y}{x}\right)$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

$\frac{d v}{d x} \cdot x+v=v-\cos ^{2} v$

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

$\sec ^{2} v d v=-\frac{d x}{x}$

Integrating both sides

$\int \sec ^{2} v d v=-\int \frac{d x}{x}$

tan v=-log|x|+C

$\tan \frac{y}{x}=C-\log |x|$

Question 28

$x \frac{d y}{d x}-y+x \sin \frac{y}{x}=0$

Sol :

$x \frac{d y}{d x}-y+x \sin \frac{y}{x}=0$

$x \frac{d y}{dx}=y-x \sin \frac{y}{x}$

$\frac{d y}{d x}=\frac{y}{x}-\sin \frac{y}{x}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{y}{x}-\sin \frac{y}{x}$

$\frac{d v}{d x}-x+v=v-\sin v$

$\frac{d v}{\sin v}=-\frac{d x}{x}$

$\operatorname{cosec}v d v=-\frac{d}{x}$

Integrating both sides

$\int \operatorname{cosec} v d v=-\int \frac{d x}{x}$

$\log \left|\tan \frac{v}{2}\right|=-\log | x|+\log | c|$

$\log \left|\tan \left(\frac{y}{2 x}\right)\right|=\log \frac{c}{x}$

$\tan \left(\frac{y}{2 x}\right)=\frac{c}{x}$

$x \tan \left(\frac{y}{2 x}\right)=C$

Question 29

अवकल समीकरण x²dy+y(x+y)dx=0 को हल करें यदि y=1 जब x=1

Sol :

x²dy+y(x+y)dx=0

x²dy=-(xy+y²)dx

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

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

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

$\frac{d v}{dx} \cdot x+v=-\frac{x^{2}\left(v+v^{2}\right)}{x^{2}}$

$\frac{d v}{d x} \cdot x=-v-v^{2}-v$

$\frac{d v}{d x} \cdot x=-v^{2}-2 v$

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

$\frac{d v}{v(v-z)}=-\frac{d x}{x}$

Integrating both sides

$\int \frac{d v}{v(v+2)}=-\int \frac{dx}{x}$

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

$\frac{1}{2}\left[\log {v}-\log \left| v+2\right|\right]=-\operatorname{log}|x|+\log c$

$\frac{1}{2} \log \frac{v}{v+2}+\log x=\log C$

$\frac{1}{2} \log \left|\frac{\frac{y}{x}}{\frac{y}{x}+2} \right|+\log x=\log C$

$\frac{1}{2}\left[\log \left| \frac{\frac{y}{x}}{\frac{y+2 x}{x}}\right|+\log x^{2}\right]=\log C$

$\log \frac{y}{y+2 x} \times x^{2}=\log C^{2}$

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

y=1 जब x=1

$\frac{1^{2} \cdot 1}{1+2(1)}=C^{2}$

$C^{2}=\frac{1}{3}$

∴अवकल समीकरण का हल

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

3x²y=y+2x

Question 30

अवकल समीकरण (x+y)dy+(x-y)dx=0 को हल करें यदि y=1 जब x=1

Sol :
(x+y)dy+(x-y)dx=0

(x+y)dy=-(x-y)dx

$\frac{d y}{d x}=\frac{(y-x)}{(x+y)}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{y-x}{x+y}$

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

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

$\frac{d v}{dx} \cdot x=\frac{v-1}{1+v}-v$

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

$\frac{d v}{dx} \cdot x=-\frac{\left(1+v^{2}\right)}{1+v}$

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

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

Integrating both sides

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

$\tan ^{-1} v+\frac{1}{2} \log |1+v^{2}|=-\log x+C$

$\frac{1}{2}\left[2 \tan ^{-1} \frac{y}{x}+\log \left|1+\frac{y^{2}}{x^{2}} \right|\right]=-\log x+C$

$2 \tan ^{-1} \frac{y}{x}+\log y\left|\frac{x^{2}+y^{2}}{x^{2}} \right|+\log y x^{2}=2C$

$2 \tan ^{-1} \frac{y}{x}+\log \left| \frac{x^{2}+y^{2}}{x^{2}} \cdot x^{2}\right|=2C$

y=1 जहाँ x=1

$2 \tan ^{-1} \frac{1}{1}+\log |1^2+1^{2}|=2C$

$2\frac{\pi}{4}+\log 2=2C$

$\frac{\pi}{2}+\log 2=2C$

अवकल समीकरण का हल

$2 \tan ^{-1} \frac{y}{x}+\log|x^2+y^2|=\frac{\pi}{2}+\log 2$

Question 31

अवकल समीकरण (x²-y²)dx+2xydy=0 को हल करें यदि y=1 जब x=1

Sol :

(x²-y²)dx+2xydy=0

2xydy=-(x²-y²)dx

$\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$..(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{dx} \cdot x+v=\frac{y^{2}-x^{2}}{2 x y}$

$\frac{d v}{dx} \cdot x+v=\frac{v^{2} x^{2}-x^{2}}{2 x \cdot v x}$

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

$\frac{d v}{d x} \cdot x=\frac{v^{2}-1}{2 v}-v$

$\frac{d v}{dx} \cdot x=\frac{v^{2}-1-2 v^{2}}{2 v}$

$\frac{d v}{d x} \cdot x=\frac{-1-v^{2}}{2 v}$

$\frac{dv}{dx} \cdot x=-\frac{\left(1+v^{2}\right)}{2 v}$

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

Integrating both sides

$\int \frac{2 v}{1+v^{2}} d v=-\int \frac{d x}{x}$

log|1+v²|=-log|x|+log|C|

$\log \left|1+\frac{y^{2}}{x^{2}}\right|+\log x=\log C$

$\log \left|\frac{x^{2}+y^{2}}{x^{2}} \times x\right|=\log C$

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

x²+y²=Cx

y=1 जब x=1

1²+1²=C(1)

2=C

∴अवकल समीकरण का हल;

x²+y²=2x

Question 32

अवकल समीकरण 2xy+y²-2x²$\frac{dy}{dx}=0$ को हल करें यदि y=2 जब x=1

Sol :

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

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

$\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

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

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

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

$\frac{2}{v^{2}} d v=\frac{d x}{x}$

Integrating both sides

$2 \int \frac{1}{v^{2}} d v=\int \frac{d x}{x}$

$-\frac{2}{v}=\log|x|+c$

$-\frac{2 x}{y}=\log |x|+C$

∵y=2 जब x=1

$\frac{-2(1)}{2}=\log|1|+C$

-1=C

∴अवकल समीकरण का हल

$-\frac{2 x}{y}=\log |x|-1$

$\frac{2 x}{y}=-\log |x|+1$

2x=y(1-log|x|)

Question 33

अवकल समीकरण $\left(x \sin ^{2} \frac{y}{x}-y\right) d x+x d y=0$ को हल करें यदि $y=\frac{\pi}{4}$ जब x=1

Sol :

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

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

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

$\frac{d y}{dx}=\frac{y}{x}-\sin ^{2} \frac{y}{x}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

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

$\frac{d v}{d x} \cdot x+v=v-\sin ^{2} v$

$\frac{d v}{\sin^{2} v}=-\frac{d x}{x}$

$-\operatorname{cosec}^{2} v dv=\frac{dx}{x}$

Integrating both sides

$-\int \operatorname{cosec}^{2} v d v=\int \frac{d x}{x}$

cot v=log |x|+C

$\cot \left(\frac{y}{x}\right)=\log |x|+C$

$y=\frac{\pi}{4}$, जब x=1

$\cot \left(\frac{\frac{\pi}{4}}{1}\right)=\log |1|+c$

1=C

∴अवकल समीकरम का हल

$\cot \left(\frac{y}{x}\right)=\log |x|+1$

Question 34

अवकल समीकरण $\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec} \frac{y}{x}=0$  को हल करें यदि y=0 जब x=1

Sol :

$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec} \frac{y}{x}=0$

$\frac{d y}{d x}=\frac{y}{x}-\text{cosec} \frac{y}{x}$...(i)

यह एक समघातीय अवकल समीकरण है।

y=vx ⇒ $v=\frac{y}{x}$

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{dv}{dx}x+v$..(ii)

समीकरण (i) तथा (ii) से,

$\frac{d v}{d x} \cdot x+v=\frac{y}{x}-\operatorname{cosec} \frac{y}{x}$

$\frac{d y}{dx} \cdot x+v=v-\text{cosec } v$

$\frac{d v}{-\operatorname{cosec} v}=\frac{dx}{x}$

$-\sin v d v=\frac{dx}{x}$

Integrating both sides

$-\int \sin v d v=\int \frac{dv}{x}$

cos v=log|x|+C

$\cos \frac{y}{x}=\log |x|+C$

∵y=0 जब x=1

$\cos \left(\frac{0}{1}\right)=\log |1|+C$

1=C

∴अवकल समीकरण का हल

$\cos \frac{y}{x}=\log |x|+1$