Non-linear partial differential equation of order 1 Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $\displaystyle p = \frac{\partial z}{\partial x}$, $\,\displaystyle q = \frac{\partial z}{\partial y}$
$f_p=2q+2x\\ f_q=2p+2x\\ f_z=0\\ f_x=2q+2x\\  f_y=2p+2y$
Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
I deduce $$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I am not sure how to proceed from here. Kinda stuck.
 A: $$2(pq+py+qx)+x^2+y^2=0$$
You got to :
$$p+q+x+y=a \\ (p-q)^2-(x-y)^2+2(p-q)(x-y)=b$$
I checked and agree.
HINT :
Solve this system on two equations for the two unknown $p(x,y)$ and $q(x,y)$. 
For example put $p=a-q-x-y$ into the second equation. It becomes a quadratic equation to be solved for $q$. 
Now you have explicitly $z_x=p(x,y)$ and $z_y=q(x,y)$. Integrate.
$z=\int p(x,y)dx+F(y)$ and $z=\int q(x,y)dy+G(x)$
Determine $F(y)$ and $G(x)$ to make them consistent.
The integrals are complicated. This seems a boring calculus. I prefer to simplify the original PDE at first place. Let :
$$z(x,y)=u(x,y)-\frac12 x^2-\frac12 y^2$$
$p=z_x=u_x-x=P-x$
$q=z_y=u_y-y=Q-y$
$$2\left((P-x)(Q-y)+(P-x)y+(Q-y)x\right)+x^2+y^2=0$$
After simplification :
$$PQ+\frac12(x-y)^2=0$$
$$u_xu_y+\frac12(x-y)^2=0$$
$$\frac{dx}{Q}=\frac{dy}{P}=\frac{du}{2PQ}=\frac{dP}{y-x}=\frac{dQ}{x-y}=ds$$
$$\begin{cases}
P+Q=c_1 \\
PQ=-\frac12(x-y)^2
\end{cases}$$
etc.
Note: $P+Q=c_1$ is equivalent to your equation $p+q+x+y=a$ .
Note: It is a pity that no boundary condition be specified.
A: Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
Adding first two and last two fractions
$$\frac{dp+dq}{2p+2y+2q+2x}=\frac{dx+dy}{-2q-2y-2p-2x}$$
This implies, $a$ being an arbitrary constant $$p+x+q+y=a \tag{1}$$
Now the equation can be rewritten as $$2(p+x)(q+y)+(x-y)^2=0\tag{2}$$
Using $1,2$ $$(p+x)-(q+y)=\sqrt{a^2-2(x-y)^2}$$
Solving for $p+x$ and $q+y$ ,
$$\begin{align}  
p+x &=  \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q+y &=  \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
This gives
$$\begin{align}  
p&= -x+ \frac12(a+\sqrt{a^2+2(x-y)^2}) \\
q &= -y+ \frac12(a-\sqrt{a^2+2(x-y)^2})
\end{align}$$
$$dz=pdx+qdy$$
Substituting and integrating gives $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y) +\frac12\int\sqrt{a^2-2(x-y)^2}d(x-y)$$
So the solution is $$z=-\frac{x^2+y^2}{2}+\frac{a}{2}(x+y)+\dfrac{a^2\ln\left(\left|a\sqrt{2(x-y)^2+a^2}+\sqrt{2}\left|a\right|(x-y)\right|\right)}{2^\frac{5}{2}}+\dfrac{a(x-y)\sqrt{\frac{2(x-y)^2}{a^2}+1}}{2^2}+b$$
