Transforming Euler's Equation to Constant Coefficients

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Homework Help Overview

The problem involves transforming Euler's equation, specifically the form x^2y'' + pxy' + qy = 0, into a format with constant coefficients using the substitution u = ln(x). Participants are exploring the implications of this transformation and the necessary derivatives involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to normalize the equation and considers the substitution of variables. Questions arise regarding the transformation of derivatives and how they relate to the coefficients in the equation.

Discussion Status

Participants are actively discussing the transformation process, with some providing insights on the derivatives involved. There is a recognition of the complexity of the second derivative and its implications for the equation, but no consensus has been reached on the correct approach.

Contextual Notes

There are indications of confusion regarding the application of the product rule in differentiation and how the derivatives relate to the coefficients in the original equation. The discussion reflects an ongoing exploration of these concepts without a clear resolution.

Telemachus
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Homework Statement


Hi. I have this problem, which says: The equation x^2y''+pxy'+qy=0 (p and q constants) is called Euler equation. Demonstrate that the change of variable u=\ln (x) transforms the equation to one at constant coefficients.

I haven't done much. I just normalized the equation: y''+\displaystyle\frac{p}{x}y'+\displaystyle\frac{q}{x^2}y=0

Then P(x)=\displaystyle\frac{p}{x} and Q(x)=\displaystyle\frac{q}{x^2}

What should I do now? I thought instead of doing x= e^u,
then y''+ \displaystyle\frac{p}{e^u}y' + \displaystyle\frac{q}{e^{2u}} y=0 may be this is the right way, cause it seems more like following the problem suggestion.
 
Last edited:
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what is d/dx in terms of d/du?
 
u=\ln (x) \rightarrow du=\frac{1}{x}dx
 
Now find out what d^2/dx^2 is in terms of d^2/dx^2 and d/du.
 
I see the relation, u=\ln (x) \rightarrow du=\frac{1}{x}dx \rightarrow d^2u=\frac{1}{x^2}d^2x
Right?
I think I got it. Thanks.
 
I thought I got it but no :P
How do I use the fact of this derivatives appears as the coefficients in the equation?
 
You've done you're second derivative wrong, use the product rule.
 
d^2u=\frac{1}{x^2}dx+\frac{1}{x}d^2x Thats it?
 

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