Question

Reduce the equation Zxx - (1+y)² Zyy = 0 to canonical form

Answer 1

from the given equation,

R=1, S=0,T=-(1+y)²

By using S²-4RT=0-4(-(1+y)² =4(1+y)² >0.

Therefore, the given equation is a hyperbolic partial Differential equation.

And its canonical form is given by

$\frac{\partial^2z}{\partial u\partial v}=0$