Solve $xdx-ydy=y^2(x^2-y^2)dy$

Question

Question:

$$xdx-ydy=y^2(x^2-y^2)dy$$

I'm having trouble matching the solution in the book, which is $\frac{1}{2}\ln(x^2-y^2)=\frac{1}{3}y^3+C$.

I'm getting an integral that requires the incomplete gamma function.

My attempt:

Rewrite the equation:

$x + (-(x^2 - y^2) y^2 - y)\frac{dy}{dx} = 0$

This is not an exact equation, but I found the integrating factor: $μ(y) = e^{-(2 y^3)/3}$

Multiply both sides of $x + \frac{dy}{dx} (-(x^2 - y^2) y^2 - y) = 0$ by $μ(y):$

$xe^{-\frac{2}{3}y^3} + (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y) \frac{dy}{dx} = 0$

Let $R(x,y) =xe^{-\frac{2}{3}y^3} $ and $S(x,y) = (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y)$. So I want to seek $f(x,y)$ such that $\frac{\partial f(x,y)}{x} = R(x,y)$ and $\frac{\partial f(x,y)}{y} = S(x,y)$

Integrating w.r.t $x$:

$f(x,y) = \int xe^{-\frac{2}{3}y^3}\,dx = \frac{1}{2}x^2 e^{-\frac{2}{3}y^3} + g(y)$

$\frac{dg(y)}{dy} = e^{-\frac{2}{3}y} y (y^3 - 1)$

Integrating w.r.t $y$:

$g(y) = \int e^{-\frac{2}{3}y} y (y^3 - 1) dy = ?$

I'm stuck here.

Answer 1

Looks easier if you try to substitute $u=x^2-y^2 \implies u'=2xx'-2y$ $$xdx-ydy=y^2(x^2-y^2)dy$$ $$xx'-y=y^2(x^2-y^2)$$ $$u'=2y^2u$$ The DE is separable

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More simply: $$xdx-ydy=y^2(x^2-y^2)dy$$ $$dx^2-dy^2=2y^2(x^2-y^2)dy$$ The DE is separable. $$\dfrac {d(x^2-y^2)}{x^2-y^2}=2y^2dy$$ Integrate. $$\ln (x^2-y^2)=\dfrac 23 y^3+C$$

Answer 2

Your way is correct. https://www.wolframalpha.com/input?i=Integral+e%5E%7B-2%2F3y%5E3%7D%28y%5E4-y%29dy

I saw this trick: $$\frac{d(x^2-y^2)}{x^2-y^2}=2y^2dy$$ $$\ln(x^2-y^2)=\frac{2}{3}y^3+2c$$