Question #172952

Solve the PDE dz/dx+dz/dy = z^2


1
Expert's answer
2021-04-29T17:57:53-0400

Solve the PDE dz/dx+dz/dy = z^2

Solution:

A parametrization invariant is:

dx=dy=dzz2dx=dy=\frac{dz}{z^2}

From dx=dudx = du ,

xy=C1x-y=C_1 which is a first characteristic equation.

From dy=dzz2dy=\frac{dz}{z^2} ,

y=1z+C2y=-\frac{1}{z}+C_2 is the second characteristic equation.

The general solution of the PDE expressed on the form of an implicit equation is :

Φ((xy),(y+1z))=0\Phi((x-y),(y+\frac1z))=0

where Φ\Phi is any differentiable function of two variables.

Answer: Φ((xy),(y+1z))=0\Phi((x-y),(y+\frac1z))=0 .


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