Question

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

• A. $Ky=e^{y^3/2x^2}$
• B. $Kx=e^{x^3/2y^3}$
• C. $Kx=e^{3x^3/y^3}$
• D. None of these
  

Answer 1

Answer: (D)

None of these

The given differential equation is a homogeneous differential equation.

Re-arranging the terms give us

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

Let $y=vx$

$v+\frac{xdv}{dx}=\frac{dy}{dx}$

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

$\frac{xdv}{dx}=\frac{v}{1+v^3}-v$

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

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

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

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

Integrating both sides :-

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

$\frac{-v^{-3}}{3}+\ln v=-lnx+C$

$\frac{-v^3}{3}+\ln vx=C$

$\ln vx=C+\frac{v^3}{3}$

Now $vx=y$

$\ln y=C+\frac{y^3}{3x^3}$

$y=e^{C+\frac{y^3}{3x^3}}$

$y=e^C.e^{\frac{y^3}{3x^3}}$

$y=K{e}^{\frac{y^3}{3x^3}}$