Question

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

• A. $-\frac{1}{3}\frac{x^3}{y^3}+\log y=C$
• B. $-\frac{1}{3}\frac{x^3}{y^3}-\log y=C$
• C. $\frac{x^3}{y^3}+\log y=C$
• D. None of these

Answer 1

The correct option is A $-\frac{1}{3}\frac{x^3}{y^3}+\log y=C$

$x^{2}ydx=(x^3+y^3)dy\Rightarrow \frac{dy}{dx}=\frac{x^{2}y}{x^3+y^3}$

Divide and multiply $R.H.S$ by $x^3$

$\frac{dy}{dx}=\frac{\frac{y}{x}}{1+\frac{y^3}{x^3}}..\left(1\right)$

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

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

$x\frac{dv}{dx}=-\frac{v^4}{1+v^3}$

$\frac{(1+v^3)dv}{v^4}=-\frac{dx}{x}$

$\frac{dv}{v^4}+\frac{dv}{v}=-\frac{dx}{x}$

Integrate both sides,

$-\frac{}{3v^3}+\log |v|=-\log x+c$

Undo $v=\frac{y}{x}$

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

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