Find the solution for the following Cauchy's problem
$y(1-y^2)\frac{\partial z}{\partial x}-(1+x-y^2x-y^2)\frac{\partial > z}{\partial y}=0, x\geq 0, y\in\mathbb{R}$
$z(x,0)=x, x\geq0$
I found the first integral: $\phi _1=x+\frac{x^2}{2}-\frac{y^2}{2}$
Could you please tell me how to continue? I'm stuck
Your characteristic equation is correct. The general solution is : $$z(x,y)=F\big(x+\frac{x^2}{2}-\frac{y^2}{2}\big)$$ $F(X)$ is an arbitrary function (to be determined according to the condition).
Condition : $z(x,0)=x=F\big(x+\frac{x^2}{2}\big)$
With $\quad X=x+\frac{x^2}{2} \quad\implies\quad x=-1\pm\sqrt{2X+1}$
$$x=F(X)=-1\pm\sqrt{2X+1}$$ Now the function $F(X)$ is known. We put it into the above general solution where $X=x+\frac{x^2}{2}-\frac{y^2}{2}$ $$z(x,y)=-1\pm\sqrt{2\big(x+\frac{x^2}{2}-\frac{y^2}{2}\big)+1}$$ $$z(x,y)=-1\pm\sqrt{(x+1)^2-y^2}$$ The sign is determined according to $z(x,0)=x$ $$\boxed{z(x,y)=-1+\sqrt{(x+1)^2-y^2}}$$