Question

Solution of the differential equation $\frac{dy}{dx}=\frac{x^{2}y}{x^3+y^3}$ is:
(where $C$ is integration constant)

• A. $|y|=C{e}^{\left(\frac{x^3}{3y^3}\right)}$
• B. $|y|=C{e}^{\left(\frac{x^2}{2y^4}\right)}$
• C. $y=C{e}^{\left(\frac{x^4}{4y^4}\right)}$
• D. $y=C{e}^{\left|\frac{x}{y}\right|}$

Answer 1

The correct option is A $|y|=C{e}^{\left(\frac{x^3}{3y^3}\right)}$

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

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

So, we have:
$\frac{dy}{dx}=v+x\frac{dv}{dx}=\frac{v}{v^3+1}$
$\Rightarrow x\frac{dv}{dx}=-\frac{v^4}{v^3+1}$
$\int \frac{(v^3+1)}{v^4}dv=-\int \frac{dx}{x}\Rightarrow \ln |v|+\frac{v^{-3}}{-3}=-\ln |x|+c$
$\Rightarrow \frac{1}{3}{\left(\frac{x}{y}\right)}^3-\ln |x|-\ln |v|+c=0$
$\therefore \ln |y|-c=\frac{x^3}{3y^3}\Rightarrow |y|=C{e}^{\frac{x^3}{3y^3}}$