Change of variables in differential equation

In summary, the conversation involves transforming an equation using variables and applying the chain rule and product rule. After some calculations, the solution is found in a book by B.P. Demidowich, which states that one of the partial derivatives is equal to 0. The individual is still unsure about the validity of their method and is seeking clarification on whether they have missed something.
  • #1
michalpp
2
0

Homework Statement



I have to transform the following equation using variables [tex](u,v,w(u,v))=(yz-x,xz-y, xy-z)[/tex]:

[tex](xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.[/tex]


Homework Equations


chain rule: [tex]
\frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{\partial du}{\partial dx} + \frac{\partial w}{\partial v} \frac{\partial dv}{\partial dx}
[/tex]


The Attempt at a Solution


Using the chain rule and the product rule:

[tex]\frac{\partial w}{\partial x}=\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}[/tex]

and a similar expression for [tex]\frac{\partial w}{\partial y}[/tex].
On the other hand

[tex]w=xy-z[/tex], so

[tex]\frac{\partial w}{\partial x}=y-\frac{\partial z}{\partial x}[/tex]

(and similar for [tex]\frac{\partial w}{\partial y}[/tex]), so

[tex]\frac{\partial z}{\partial x}(y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v})-\frac{\partial w}{\partial u}+z\frac{\partial w}{\partial v}=y-\frac{\partial z}{\partial x}[/tex] and therefore:

[tex]\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}[/tex] and

[tex]\frac{\partial z}{\partial y}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=x+\frac{\partial w}{\partial v}-z\frac{\partial w}{\partial u}.[/tex]

After multiplying the given equation by

[tex][y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1][/tex]

and writing

[tex]\frac{\partial z}{\partial x}[y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1][/tex] as [tex]y+\frac{\partial w}{\partial u}-z\frac{\partial w}{\partial v}[/tex] I get

[tex]\frac{\partial w}{\partial v}(1-x^2-y^2-z^2-2xyz)=0.[/tex]

If
[tex][y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]\neq 0[/tex]

this equation is equivalent to the given one. But what if [tex][y\frac{\partial w}{\partial u}+x\frac{\partial w}{\partial v}+1]=0[/tex]?.

I'm kind of stuck at this point, cause in the first case

[tex]\frac{\partial w}{\partial v}=0[/tex] or [tex](1-x^2-y^2-z^2-2xyz)=0[/tex].

I was thinking of a way to change [tex](x,y,z)[/tex] to [tex](u,v,w)[/tex] in the last equation, but without success. I also have no idea what to do in the second case. So what should I do now? Is this a correct method to solve a problem like this?
 
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  • #2
I've found the answer to this problem in a book by B.P. Demidowich - it says
[tex]
\frac{\partial w}{\partial v}=0
[/tex]
So this means that I need a proof that
[tex]
(1-x^2-y^2-z^2-2xyz)\neq 0
[/tex], or do I still miss something?

[tex]
(1-x^2-y^2-z^2-2xyz)=(xy+z)z+(1-y^2)(-1)+x(x+yz)
[/tex]
looks similar to
[tex]
(xy+z)\frac{\partial z}{\partial x}+(1-y^2)\frac{\partial z}{\partial y}=x+yz.
[/tex] but I don't have other ideas.
 
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FAQ: Change of variables in differential equation

What is a change of variables in a differential equation?

A change of variables in a differential equation refers to the process of substituting one variable for another in order to simplify or solve the equation. This can be done to make the equation easier to work with, or to transform it into a form that is more familiar or easy to solve.

Why is a change of variables useful in solving differential equations?

A change of variables can be useful in solving differential equations because it can transform a complex equation into a simpler one, making it easier to find a solution. It can also help to identify patterns or relationships within the equation that may not have been apparent before.

What types of changes can be made to variables in a differential equation?

There are several types of changes that can be made to variables in a differential equation, including substitution, transformation, and integration by parts. These changes can be used to simplify the equation, eliminate certain terms, or convert the equation into a different form.

When should a change of variables be used in solving a differential equation?

A change of variables should be used when the original equation is too complex or difficult to solve directly. It can also be used when the equation contains multiple variables, and it would be easier to work with a single variable. Additionally, a change of variables can be used to transform an equation into a more familiar form for which a solution is known.

What are some common techniques for performing a change of variables in a differential equation?

Some common techniques for performing a change of variables in a differential equation include substitution, integration by parts, and the method of undetermined coefficients. These techniques involve manipulating the variables in the equation in a strategic way to simplify or transform it into a more manageable form.

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