Method of Characteristics for Solving Partial Differential Equations

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The discussion focuses on solving a partial differential equation using the method of characteristics. The user is attempting to parameterize the dependent variables and has derived a system of autonomous equations but is struggling with the equation for z. They reference a handout that provides a solution, specifically how to manipulate the dz/dr equation to match the provided form. The user seeks clarification on the steps taken in the handout, particularly how to transition from their derived equation to the one presented. The conversation emphasizes the need for detailed working to understand the solution process better.
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Homework Statement


I am trying to solve the following equation using the method of characteristics:

∂u/∂x + (xy)(∂u/∂y) + 2x2zLog[y](∂u/∂z) = 0

I'm really just trying to follow along the solution provided here:
http://www.ucl.ac.uk/~ucahhwi/LTCC/sectionA-firstorder.pdf

on page 9.

The Attempt at a Solution



I get that we have to parameterize the dependent variables so x=x(r), y=y(r), z=z(r). Then demand that du/dr=0, from which we get a system of three autonomous equations:

dx/dr = 1; dy/dr = xy; dz/dr = 2x2Log[y].

Then we have to solve these to get the form of u. My trouble is with solving the z equation. According to the handout I linked to above the right hand side of the dz/dr equation becomes:

2x2zLog[y] = -r4zLog[y0], where r=x and y=y0Exp[r^2/2] (these expressions for x and y come from solving the other two equations).

I don't understand how we can make this step. When I plug in for y and x to the left side of the above I come up with y=2zr^2Log[y0] + r^4/2 ...

Can somebody tell me what I'm doing wrong and how to get the answer given in the handout?

Thank you!
 
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Sturk200 said:
I come up with y=2zr2Log[y0] + r4/2...
I don't see how you get that. Please post your working.
 
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