# Diff.equation transformation by change of variables

• TheMercury79
In summary, the assignment was to transform a given differential equation by changing the variables. The resulting expressions for the partials were used to obtain the transformation, but there was some uncertainty about the correctness of the solution due to the non-continuity of certain variables and their derivatives. After taking second partials directly without variables, a new transformation was obtained.
TheMercury79

## Homework Statement

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##

## Homework Equations

Implicit differentiation in multivariable calculus

## 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 continuous 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.

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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:

## 1. What is a differential equation?

A differential equation is a mathematical equation that describes how a variable changes over time, based on the rate at which the variable is changing. It can involve one or more variables and their derivatives.

## 2. What is a change of variables in a differential equation?

A change of variables in a differential equation is a transformation that involves replacing the existing variables with new ones in order to simplify the equation or make it easier to solve. This is often done to eliminate terms or make the equation more manageable.

## 3. Why is change of variables useful in solving differential equations?

Change of variables can be useful in solving differential equations because it can help to simplify the equation, making it easier to solve. It can also help to transform the equation into a more familiar form, making it easier to apply known solution techniques.

## 4. What are some common examples of change of variables in differential equations?

Some common examples of change of variables in differential equations include substitution, transformation to polar or spherical coordinates, and using the chain rule to simplify derivatives.

## 5. Are there any limitations to using change of variables in differential equations?

Yes, there are limitations to using change of variables in differential equations. In some cases, the transformed equation may be more complex or difficult to solve than the original equation. It is also important to ensure that the transformation is valid and does not introduce any new solutions or change the nature of the problem being solved.

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