Solve the characteristic of the PDE $z=p^2-q^2$, and find the integral surface which passes through the parabola $4z+x^2=0,y=0$
I used Clairaut's methods and found a relation between $p,q$ as $\frac{q}{p}=c(constant),q=pc$
Thus $z=p^2(1-c^2)\;$, and $\;\;dz=pdx+qdy=p(dx+cdy)$. Substituting the value of $p$, we get $$dz=\frac{\sqrt {z}}{\sqrt {1-c^2}}(dx+cdy)$$
Using this I solved the differential equation and got the solution as $$2\sqrt z=\frac{x}{\sqrt {1-c^2}}+\frac{c}{\sqrt {1-c^2}}y+c_1$$, where $c,c_1$ are constants. Now I am not able to find the integral surface which passes through the given parabola. Please help solve the problem
$2\sqrt z=\frac{x}{\sqrt {1-c^2}}+\frac{c}{\sqrt {1-c^2}}y+c_1$ is correct. $$z=\frac14\left(\frac{x+cy}{\sqrt {1-c^2}}+c_1\right)^2\tag 1$$
$$z(x,0)=\frac14\left(\frac{x}{\sqrt {1-c^2}}+c_1\right)^2$$ $$4z(x,0)+x^2=\left(\frac{x}{\sqrt {1-c^2}}+c_1 \right)^2+x^2=0$$ $$\left(\frac{1}{(1-c^2)}+1 \right)x^2+\frac{c_1}{\sqrt{1-c^2}}x+c_1^2=0$$ $$\begin{cases} c_1=0\\ \frac{1}{(1-c^2)}+1=0 \quad\implies\quad c=\pm\sqrt{2} \end{cases}$$ We put $c_1=0$ and $c=\pm\sqrt{2}$ into Eq.$(1)$ : $$z=\frac14\left(\frac{x\pm\sqrt{2}\:y}{i}\right)^2$$ $$z(x,y)=-\frac14\left(x\pm\sqrt{2}\:y\right)^2$$ One can check that this solution satisfies both the PDE and the boundary condition.