Solve partial differential equation $$F = zp(x+y)+q^2-pq-z^2=0, p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}$$
My attempt:
Differentiating partially wrto x,y,z,p,q
$F_x = zp-q \\ F_y=zp , \\F_p = z(x+y) \\ F_q = 2q-p \\ F_z=-2z+px+py \\ \frac{dx}{-F_p}=\frac{dy}{-F_q}=\frac{dz}{-pF_p-qF_q}=\frac{dp}{F_x+pF_z}=\frac{dq}{F_x+qF_z}$
Charpit's equations gives:
$$\frac{dx}{-z(x+y)} = \frac{dy}{p-2q} = \frac{dp}{p^2(x+y)-pz-q} = \frac{dq}{zp-q(p-2q)} = \frac{dz}{(p-2q)q-pz(x+y)} $$
Taking $dx, dy, dz$ terms, i got $$dz=pdx+qdy$$
How to proceed further?
Pls find the detailed calculation part in following image
In order to simplify the Lagrange-Charpit equations, let's make the following changes of variables:
$z=e^u$; then $(p,q)=(z_x,z_y)=(u_xe^u, u_ye^u)$ and the PDE $$ zp(x+y)+q^2-pq-z^2=0 \tag{1} $$ becomes $$ u_x(x+y)+u_y^2-u_xu_y-1=0. \tag{2} $$
$u(x,y)=v(\xi,\eta)$, where $(\xi, \eta)=(x+y, x-y)$; then $u_x=v_{\xi}\xi_x+v_{\eta}\eta_x=v_{\xi}+v_{\eta}=:P+Q$ and, similarly, $u_y=P-Q$. In terms of the new variables, Eq. $(2)$ becomes $$ (P+Q)\xi-2PQ+2Q^2-1=0. \tag{3} $$
The Lagrange-Charpit equations for the PDE $(3)$ are $$ \frac{d\xi}{\xi-2Q}=\frac{d\eta}{\xi-2P+4Q}=\frac{dP}{-P-Q}=\frac{dQ}{0}=\frac{dv}{(P+Q)\xi-4PQ+4Q^2}. \tag{4} $$ The null denominator in $\frac{dQ}{0}$ implies that $v_{\eta}=Q=a,$ hence $v=a\eta+f(\xi)$ and $P=v_{\xi}=f'(\xi).$
Substituting $P$ and $Q$ in $(3)$, we conclude that $f$ must satisfy the ODE $$ (f'+a)\xi-2af'+2a^2-1=0 \implies f'=-a+\frac{1-4a^2}{\xi-2a}, \tag{5} $$ whose solution is $$ f(\xi)=-a\xi+(1-4a^2)\ln|\xi-2a|+b. \tag{6} $$ Therefore, $$ v(\xi,\eta)=a\eta+f(\xi)=a(\eta-\xi)+(1-4a^2)\ln|\xi-2a|+b $$ $$ \implies u(x,y)=-2ay+(1-4a^2)\ln|x+y-2a|+b $$ $$ \implies z(x,y)=e^u=ce^{-2ay}|x+y-2a|^{1-4a^2} \tag{7} $$ is a complete integral of the PDE $(1)$.