Let $p= \frac{\partial z}{\partial x}, ~q= \frac{\partial z}{\partial y}$. Find the general solution of the partial differential equation $z = p x+ qy +p+q -pq$, by finding the envelope of those planes that pass through the origin. It is given that, $z =ax+ by + a+b -ab$ is a complete integral. (This question is part of a problem from the book "Elements of partial differential equations" by Ian N. Sneddon.)
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Hint:
$z=x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}+\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}-\dfrac{\partial z}{\partial x}\dfrac{\partial z}{\partial y}$
$z=(x+1)\dfrac{\partial z}{\partial x}+(y+1)\dfrac{\partial z}{\partial y}-\dfrac{\partial z}{\partial x}\dfrac{\partial z}{\partial y}$
Let $\begin{cases}r=x+1\\s=y+1\end{cases}$ ,
Then $z=r\dfrac{\partial z}{\partial r}+s\dfrac{\partial z}{\partial s}-\dfrac{\partial z}{\partial r}\dfrac{\partial z}{\partial s}$
doraemonpaul
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The planes $z=ax+by+a+b-ab$ pass through the origin if $a$ and $b$ satisfy the relation $$ a+b-ab=0 \implies b=\frac{a}{a-1}. \tag{1} $$ Substituting $(1)$ in the equation of the planes we find $$ z=ax+\frac{ay}{a-1}. \tag{2} $$ The envelope of these planes is obtained by eliminating $a$ between $(2)$ and the equation $$ \frac{\partial}{\partial a}\left(-z+ax+\frac{ay}{a-1}\right)=0 \implies x+\frac{y}{a-1}-\frac{ay}{(a-1)^2}=0. \tag{3} $$ Solving $(3)$ for $a$ we obtain $$ a=1\pm\sqrt{\frac{y}{x}}. \tag{4} $$ Substituting $(4)$ in $(2)$ we finally obtain $$ z=\left(\sqrt{x}\pm\sqrt{y}\right)^2, \quad \text{or} \quad (z-x-y)^2=4xy. \tag{5} $$
Gonçalo
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