Question

The general solution of the differential equation $x^{2} y \,d x-\left(x^{3}+y^{3}\right) dy=0$ is

• A. $y^{3}=3 x^{3} \log (c x)$
• B. $c\left(x^{3}-y^{3}\right)=x^{2}$
• C. $\log |y|-\frac{x^{3}}{2 x^{3}}=c$
• D. $y^{2}-x^{2}=c^{2}\left(y^{2}-x^{2}\right)$

Answer 1

Answer: (C)

We have,
$x^{2} y d x-\left(x^{3}+y^{3}\right) d y=0$
$ \Rightarrow \, \frac{d y}{d x}=\frac{x^{2} y}{x^{3}+y^{3}}$
Given differentiate equation is in the form of homogeneous differentiate equation.
So, let $y=v x$
$\Rightarrow \, \frac{d y}{d x}-v+x \frac{d v}{d x} $
$\therefore \, v+x \frac{d v}{d x}=\frac{v}{1+v^{3}}$
$\Rightarrow \,x \frac{d v}{d x}=\frac{v}{1+v^{3}}-v$
$ \Rightarrow \,x \frac{d v}{d x}=\frac{v-v-v^{4}}{1+v^{3}}$
$\Rightarrow \, x \frac{d v}{d x}=\frac{-v^{4}}{1+v^{3}} $
$\Rightarrow \, \frac{1+v^{3}}{v^{4}} d v=-\frac{d x}{x}$
Integrating both side, we get
$\int\left(\frac{ l }{v^{4}}+\frac{ l }{v}\right) d v=-\int \frac{1}{x} d x$
$ \Rightarrow -\frac{1 }{3 v^{3}}+\log |v|=-\log x+c$
$\Rightarrow \, -\frac{x^{3}}{3 y^{3}}+\log \left|\frac{y}{x}\right|=-\log |x|+c $
$\Rightarrow -\frac{x^{3}}{3 y^{3}}+\log |y|=c$