PDE question that hard to see second first integral

Question

Solve the IVP. $x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y} = z- x^2-y^2, \; z\vert_{y=-2} = x-x^2 $

$\underline{\text{My Attempt:}}$

I first wrote Characteristic Equation for the correpsonding PDE. Here is the Characteristic equation:

$$\dfrac{\mathrm{dx}}{x}=\dfrac{\mathrm{dy}}{y}=\dfrac{\mathrm{dz}}{z-x^2-y^2}$$

And clearly the first, first integral is:

$$\dfrac{x}{y}=\phi_1(x,y,z)$$

But I couldn't find the second first integral. Could someone help?

Thanks in advance!

Answer 1

$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z-x^2-y^2}=\frac{2x\,dx+2y\,dy+dz}{2x^2+2y^2+(z-x^2-y^2)} = \frac{d(x^2+y^2+z)}{x^2+y^2+z}$$

$$\ln|x|=\ln|x^2+y^2+z|+\text{constant}$$ A second integral : $$\frac{x^2+y^2+z}{x}=\phi_2$$ I suppose that you can take it from here.

Answer 2

I think if you define $u(x,y) = z(x,y)+x^2+y^2$ the equation reduces to $xu_x+yu_y=u$, which is easier to analyse.

If $z=u-x^2-y^2$, then $xz_x +yz_y=xu_x-2x^2+yu_y-2y^2$, but $z-x^2-y^2=u-2x^2-2y^2$, giving the required equation.