Question

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

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

Answer 1

Answer: (C)

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