# Finding the integral surface passing through a given curve

The question here is to solve $x^2p+y^2q+z^2=0$ passing through a given curve $\Gamma : xy=x+y,z=1$

I am trying to solve it by Lagrange's method, where $\frac{dx}{x^2}=\frac{dy}{y^2}$, thus I get $1/y-1/x =c$ and then I substituted $x+y$ in place of $xy$ in denominator.

Thus I found: $(x-y)/(x+y)=c$

Is it correct ?

Soham

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You are looking for $z$ as a function of $(x,y)$. The idea is to think of the graph of $z(x,y)$ as built up from curves starting from $\Gamma$. Say we parametrize the curves as $(x(s,x_0),y(s,x_0),z(s,x_0))$ which start from $\Gamma$ at $s=0$, at the point $(x_0,y_0,z_0)$, where $z_0 = 1$ and $x_0 y_0 = x_0+y_0$ i.e. $1/x_0+1/y_0 = 1$. Then take the PDE to be a case of the chain rule $$z' = z_x x'+z_y y' = x^2z_x+y^2z_y=-z^2$$ where we take $x'=x^2$, $y'=y^2$, and $z' = -z^2$ with initial values as above. Solve these by separation of variables, integrating $x^{-2}x'=1$ and so forth, as $$\frac{1}{x}-\frac{1}{x_0}=-s, \ \ \frac{1}{y}-\Big(1-\frac{1}{x_0}\Big)=-s, \ \ \frac{1}{z}-1=s.$$ Add the first two to eliminate $x_0$, giving $1/x+1/y-1=-2s$. Then this is $-2\big(1/z-1\big)$, so you find $$z(x,y)=\frac{1}{\frac{3}{2}-\frac{1}{2y}-\frac{1}{2x}}$$This is a solution if you check it.