Question

Find the integral surface of the pde pq=z containing the curve x-0, z-y²

Answer 1

$F(z,p,q)=pq-z=0$

Trivial solution:

$z=f(x+ay)$

Let $x+ay=u$

then:

$z=f(u), \frac{\partial u}{\partial x}=1, \frac{\partial u}{\partial y}=a$

$p=\frac{\partial z}{\partial x}=\frac{dz}{du}\frac{\partial u}{\partial x}=\frac{dz}{du}$

$q=\frac{\partial z}{\partial y}=\frac{dz}{du}\frac{\partial u}{\partial y}=a\frac{dz}{du}$

Substituing values of $p$ and $q$ in the initial equation, we get:

$a(\frac{dz}{du})^2=z$

$\frac{dz}{du}=\sqrt{z/a}$

$du=dz/\sqrt{z/a}$

$u=2\sqrt{z/a}+b$

$x+ay=2\sqrt{z/a}+b$

If $x=0, z=y^2$ , then the solution is:

$ay=2y/\sqrt{a}+b$

$y=\frac{b}{a-2/\sqrt{a}}, z=(\frac{b}{a-2/\sqrt{a}})^2$

Also:

$\frac{dz}{du}=-\sqrt{z/a}$

$x+ay=-2\sqrt{z/a}+b$

$ay=-2y/\sqrt{a}+b$

$y=\frac{b}{a+2/\sqrt{a}}, z=(\frac{b}{a+2/\sqrt{a}})^2$