Answer to Question #339634 in Differential Equations for Mudasir

Question #339634

Find canonical form of Zxx +Zyy=y

1
Expert's answer
2022-05-12T18:26:24-0400

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

B24AC=04x1x1=4<0B^2-4AC=0-4x1x1=-4<0

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


And its canonical form is given by

δ2zδuδv=y\frac{\delta^2z}{\delta u \delta v}=y



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