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

Question

Solve : $x^3 \frac{dy}{dx} = y^3 +y^2\sqrt{y^2 - x^2}$

My try :

But it is not integrable

Answer 1

$$x^3\frac{dy}{dx}=y^3+y^2\sqrt{y^2-x^2}$$ let $\frac{y}{x}=v$ so $\frac{dy}{dx}=v+x\frac{dv}{dx}$

so: $$\int\frac{1}{x}dx=\int\frac{1}{v^3+v^2\sqrt{v^2-1}-v}dv$$ now using $v=\sec(u)$ we get: $$\ln|x|+C=\int\frac{\tan(u)\sec(u)}{\sec^2(u)+\sec(u)\tan(u)-\sec(u)}du$$