Question

Solve the differential equation :
$(x^3+y^3)dy-x^{2}ydx=0$

Answer 1

To Solve the differential Equation

$(x^3+y^3)dy-x^{2}ydx=0$

$(x^3+y^3)dy=x^{2}ydx$

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

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

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

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

$x=vy$

On Differentiating wrt y , we get

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

On Putting both values we get

$v+y\frac{dv}{dy}=v+\frac{1}{v^2}$

$y\frac{dv}{dy}=\frac{1}{v^2}$

$v^{2}dv=\frac{dy}{y}$

$\int v^{2}dv=\int \frac{dy}{y}$

On integrating this , we get

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

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

Where C is the Integration Constant