Find canonical form of Zxx +Zyy=y
Comparing the given equation with the equation
Azxx+Bzxy+Czyy+Dzx+Ezy+Fz=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"
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