$x(y^2+u)u_x-y(x^2+u)u_y=(x^2-y^2)u$ with $u(x,-x)=1$ for $x>0$

Question

Not homework but exercices I like to do...

Solve $x(y^2+u)u_x-y(x^2+u)u_y=(x^2-y^2)u$ with $u(x,-x)=1\;x>0$

Manipulating the auxiliar system of ODEs $\dfrac{dx}{x(y^2+u)}=\dfrac{dy}{-y(x^2+u)}=\dfrac{du}{(x^2-y^2)u}$

I get a pretty simplified expression $(x+y)d(x^2+y^2)=2du$ that I see unusable to find the general solution. I found no better expression. I know the solution has a not very complicated closed form, but I am stuck... Help appreciated.

Answer 1

Hint: Left details to you: $$\dfrac{x\ dx}{x^2y^2+x^2u}=\dfrac{y\ dy}{-x^2y^2-uy^2}=\dfrac{du}{(x^2-y^2)u}$$ $$\dfrac{x\ dx+y\ dy}{x^2y^2+x^2u-x^2y^2-uy^2}=\dfrac{du}{(x^2-y^2)u}$$ $$\color{blue}{d(x^2+y^2)=2du}$$ $$\dfrac{y\ dx}{xy(y^2+u)}=\dfrac{x\ dy}{-xy(x^2+u)}=\dfrac{du}{(x^2-y^2)u}$$ $$\dfrac{y\ dx + x\ dy}{xy(y^2-x^2)}=\dfrac{du}{(x^2-y^2)u}$$ $$\color{blue}{\dfrac{d(xy)}{xy}=-\dfrac{du}{u}}$$

Answer 2

This is a Lagrange method, assuming you know how to do it. I'll give you a hint to find $\varphi$ and then you can similarly find $\psi$, check the Jacobian and solve it. $$ \frac{\lambda dx + \mu dy + \nu du}{\lambda x(y^2+u)-\mu y(x^2+u)+\nu u(x^2-y^2)} = \frac{dx}{x(y^2+u)}=\frac{dy}{-y(x^2+u)}=\frac{du}{u(x^2-y^2)}$$ For $\lambda=x$, $\mu=y$ and $\nu=-1$ you get $$ \varphi_x=x \quad \varphi_y=y \quad \varphi_u=-1 $$ and solve it to get $\varphi(x,y,u)=\frac{x^2}{2}+\frac{y^2}{2}-u$.