# Diff.equation transformation by change of variables

#### TheMercury79

1. The problem statement, all variables and given/known data

The assignment is to transform the following differential equation: $x^2\frac {\partial^2 z} {\partial x^2}-2xy\frac {\partial^2 z} {\partial x\partial y}+y^2\frac {\partial^2 z} {\partial y^2}=0$
by changing the variables: $u=xy~~~~~~y=\frac 1 v$

2. Relevant equations
Implicit differentiation in multivariable calculus

3. The attempt at a solution
I have added a photo showing the steps of how I got to a solution. But to sum up I ended up with these
expressions for the partials:$$x^2\frac {\partial^2 z} {\partial x^2}=u^2\frac {\partial^2 z} {\partial u^2}+v^2\frac {\partial^2 z} {\partial v^2}$$$$2xy\frac {\partial^2 z} {\partial x\partial y}=2u\frac {\partial z} {\partial u}+2u^2\frac {\partial^2 z} {\partial u^2}-4v\frac {\partial z} {\partial v}-2v^2\frac {\partial^2 z} {\partial v^2}$$$$y^2\frac {\partial^2 z} {\partial y^2}=u\frac {\partial z} {\partial u}+u^2\frac {\partial^2 z} {\partial u^2}+2v\frac {\partial z} {\partial v}+v^2\frac {\partial^2 z} {\partial v^2}$$

Putting them together gives the transformation:$~~~~~4v^2\frac {\partial^2 z} {\partial v^2}+6v\frac {\partial z} {\partial v}-u\frac {\partial z} {\partial u}=0$

I'm feeling a little insecure about this because I expected the second partials to vanish completely, but the second partial with respect to v did not. And furthermore, I assumed equality of mixed partials, so when I also, just to be certain, compared the two mixed partials $\frac {\partial^2 z} {\partial x\partial y}$ and $\frac {\partial^2 z} {\partial y\partial x}$, I found out they are not equal, which also worries me a bit. But then again $\frac 1 y$ or $\frac 1 v$ and their derivatives are not continous everywhere so it might not be a problem? I'm just not sure if I have done the whole thing right and appreciate any feedback on this. See the attached photo below for a detailed solution attempt.

Thanks in advance. #### Attachments

• 72.4 KB Views: 91
Related Calculus and Beyond Homework News on Phys.org

#### TheMercury79

I think there's is a problem in how I have used the product rule when taking second partials.
Tried again by taking second partials directly without variables and arrived at:
$$2v^2\frac {\partial^2 z} {\partial v^2}+v\frac {\partial z} {\partial v}-u\frac {\partial z} {\partial u}=0$$

Which I'm somewhat more satisfied with, but wouldn't mind if anyone confirms it.

Last edited:

### Want to reply to this thread?

"Diff.equation transformation by change of variables"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving