Transforming Euler's Equation to Constant Coefficients

[Telemachus]

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.

 

[hunt_mat]

what is d/dx in terms of d/du?

 

[Telemachus]

$$u=\ln (x) \rightarrow du=\frac{1}{x}dx$$

 

[hunt_mat]

Now find out what d^2/dx^2 is in terms of d^2/dx^2 and d/du.

 

[Telemachus]

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.

 

[Telemachus]

I thought I got it but no :P
How do I use the fact of this derivatives appears as the coefficients in the equation?

 

[hunt_mat]

You've done you're second derivative wrong, use the product rule.

 

[Telemachus]

$$d^2u=\frac{1}{x^2}dx+\frac{1}{x}d^2x$$ Thats it?