Question

Find canonical form of Zxx +Zyy=y

Answer 1

Comparing the given equation with the equation

Az_xx+Bz_xy+Cz_yy+D_zx​+E_zy+F_z=G

we have

A=1, B=0, C=1, D=0, E=0, F=0, G=0A=1,B=0,C=x,D=0,E=0,F=0,G=y

The discriminant is

$B^2-4AC=0-4x1x1=-4<0$

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

And its canonical form is given by

$\frac{\delta^2z}{\delta u \delta v}=y$