Integral surface of linear PDE $xp + yq = z$ Find the integral surface of the linear partial differential equation :$$xp+ yq = z$$ which contains the circle defined by $x^2 + y^2 + z^2 = 4$, $x + y + z = 2$.
NOTE: $ p = dz/dx ;  q = dz/dy$
On solving the PDE using lagrange subsidiary equations, i got $x/y = c_1$ and $y/z = c_2$. Now, how do i find the integral surface?
 A: I am not very familiar with the "Lagrange subsidiary equations", so I shall use Charpit's method instead, if that helps.
You can write the PDE in the form $F(x,y,z,p,q)=0$ where in this case
$$F(x,y,z,p,q) = xp+yq-z \, ,  \qquad p = \frac{\partial z}{\partial x} \qquad q = \frac{\partial z}{\partial y}$$
Charpit's equations are
$$\frac{dx}{dt} = \frac{\partial F}{\partial p} \qquad \frac{dy}{dt} = \frac{\partial F}{\partial q} \qquad \frac{dz}{dt} = p\frac{\partial F}{\partial p} + q\frac{\partial F}{\partial q} \\ \frac{dp}{dt}=-\frac{\partial F}{\partial x}-p\frac{\partial F}{\partial z} \qquad \qquad \frac{dq}{dt}=-\frac{\partial F}{\partial y}-q\frac{\partial F}{\partial z}$$
where $t$ is a parameter that we will use to parameterise the solution surface. Plugging in the expression for $F$, we find that
$$\frac{dx}{dt} = x \qquad \frac{dy}{dt} = y \qquad \frac{dz}{dt} = xp+yq \qquad \frac{dp}{dt} = 0 \qquad \frac{dq}{dt} = 0$$
Solving, we get
$$x=Ae^t \qquad y=Be^t \qquad z=(AD+BE)e^t+C \qquad p=D \qquad q=E$$
where $A,B,C,D,E$ are constants.
Now we turn to the initial data. We may parameterise the given circle as
\begin{align}
x & = \frac 23 + \frac 23 \cos(s)+\frac {2}{\sqrt 3}\sin(s) \\
y & = \frac 23 + \frac 23 \cos(s)-\frac {2}{\sqrt 3}\sin(s) \\
z & = \frac 23 - \frac 43 \cos(s)
\end{align}
where $s \in [0,2\pi]$. Without loss of generality, we may assume that this is satisfied at $t=0$. Furthermore, the compatibility of Charpit's equations require that
$$F=0 \qquad \qquad \frac{dz}{ds} = p\frac{dx}{ds} + q\frac{dy}{ds} \qquad \qquad \text{at }t=0$$
Solving, we find that
\begin{align}
A & = \frac 23 + \frac 23 \cos(s)+\frac {2}{\sqrt 3}\sin(s) \\
B & = \frac 23 + \frac 23 \cos(s)-\frac {2}{\sqrt 3}\sin(s) \\
C & = 0 \\
AD+BE & = \frac 23 - \frac 43 \cos(s)
\end{align}
Thus, we find that the parametric solution is
\begin{align}
x & = \bigg(\frac 23 + \frac 23 \cos(s)+\frac {2}{\sqrt 3}\sin(s)\bigg)e^t \\
y & = \bigg(\frac 23 + \frac 23 \cos(s)-\frac {2}{\sqrt 3}\sin(s)\bigg)e^t \\
z & = \bigg(\frac 23 - \frac 43 \cos(s)\bigg)e^t
\end{align}
After a bit of algebra, the closed form solution is then found to be
$$z = -\frac{xy}{x+y}$$
