Solving the initial value problem for PDE Let consider the initial value problem given by $$xu_y-yu_x=u\quad\text{with}\ u(x,0)=h(x).$$ The answer that my teacher put on our exercises is $u=h(\sqrt{x^2+y^2})e^{\arctan(y/x)}$.

How I get the answer? I made some computations here and I find $$u=h\left(\frac{x^2-y^2}{x}\right)e^{y/x}.$$ Probably I'm wrong, but I can't see how to obtain the desired answer. Appreciate any help! :D
Remark: If someome wants to see my computations, I can put it here later.
General solution of pde is
$$u=F(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
Then solution of initial value problem is
$$u=h(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
In polar coordinates $$x=r\cos\phi,\quad y=r\sin\phi$$ pde is
$$\frac{\partial u(r,\phi)}{\partial\phi}=u(r,\phi)$$
with solution
$$u=F(r)e^\phi$$
If substitute $r=\sqrt{x^2+y^2}$, $\phi=\arctan\frac{y}{x}$ we get general solution.
General solution of pde is
$$u=F(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
Then solution of initial value problem is
$$u=h(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
In polar coordinates $$x=r\cos\phi,\quad y=r\sin\phi$$ pde is
$$\frac{\partial u(r,\phi)}{\partial\phi}=u(r,\phi)$$
with solution
$$u=F(r)e^\phi$$
If substitute $r=\sqrt{x^2+y^2}$, $\phi=\arctan\frac{y}{x}$ we get general solution.