Finding general solution to Euler equation via variation of parameters

[s3a]

Homework Statement

The problem is attached as TheProblemAndSolution.png, and everything is typewritten, so it should be easily legible (but you will likely need to zoom into read the text since the image's height is significantly larger than its width).

Homework Equations

Differential operator method.
Wronskian.
Variation of parameters.
Integration.

The Attempt at a Solution

What I understand:
1. I understand what an Euler equation is.

2. The Cramer's rule part.

3. The Wronskian part.

4. The variation of parameters part.

5. The integrating part.

6. The part about forming the general solution given the previous data.

What I don't understand:
1. Why is x^4 ln x in the question? Using WolframAlpha.com to confirm, the general solution presented at the end of the solution of this problem is that of x^2 y'' – 4x y' + 6y = x^4 sin x and not x^2 y'' – 4x y' + 6y = x^4 ln x.

2. I don't get the phi(t) = y(e^t) part.

3. I don't get the [D(D – 1) – 4D + 6] phi(t) = [(D – 3)(D – 2)] phi(t) = 0 part.

4. I don't get the y_1(x) = |x|^3 and y_2(x) = |x|^2 part.


Any input helping me fully understand this problem would be GREATLY appreciated!

 

[LCKurtz]

Your image is too small for me to read it. Nevertheless, I can help you with the $e^t$ thing. Euler equations can be transformed into a related constant coefficient equation with that substitution. A good exposition of that is here:
http://www.math.dartmouth.edu/archive/m23s06/public_html/handouts/euler_eqns.pdf

Have a look at that.

 

[s3a]

Thanks, we can now "cross out" #2. and #4. off my list of things I don't understand.

I'm still confused about the rest, though.

About the image, you can press Ctrl and scroll up, or Ctrl and the + sign, and your browser should zoom in, but alternatively, here (in the attachment of this post) is the image I uploaded in my opening post, converted to a pdf. Is this better for you?

 

[LCKurtz]

It is really hard to discuss problems when they are posted in separate attachments because we can't edit and refer to particular equations. That plus the fact it is hard to know what is really confusing you. As far as the $x^4\ln x$ versus $x^4\sin x$ is concerned, the problem has a right side of $b(x) = x^4\sin x$ or $x^4 \ln x$. Apparently the author wants you to work it with two different NH terms (two different problems), and he has only worked one of them.

 

[s3a]

I had a feeling that that's what was going on (for #1 of the list of what I don't get), but I wasn't sure if it was just something fancy I didn't understand, and I didn't want to overlook it.

So, now, I can cross out #1 from my list of what I don't get, which leaves only #3!

I'd say we're doing pretty good (with knocking out things that are confusing me). ;)

Here's the last thing remaining that confuses me.:

3. I don't get the [D(D – 1) – 4D + 6] phi(t) = [(D – 3)(D – 2)] phi(t) = 0 part.

Put differently, how do I got from substituting x with e^t to the stuff with the differential operator?

 

[LCKurtz]

I had a feeling that that's what was going on (for #1 of the list of what I don't get), but I wasn't sure if it was just something fancy I didn't understand, and I didn't want to overlook it.

So, now, I can cross out #1 from my list of what I don't get, which leaves only #3!

I'd say we're doing pretty good (with knocking out things that are confusing me). ;)

Here's the last thing remaining that confuses me.:


Put differently, how do I got from substituting x with e^t to the stuff with the differential operator?

Look again at the reference I gave in post #2. In particular, look at equations (3) and (4). If we let $\mathcal D = \frac d {dt}$ and $D = \frac d {dx}$, equation (3) becomes $t\mathcal D = D$ and equation (4) becomes $t^2\mathcal D^2 = D^2-D = D(D-1)$. Once you understand how the substitution works, you can quickly change the Euler equation expressed with $\mathcal D's$ to a constant coefficient expressed with $D's$.

Here's how you would do it with his example$$
t^2y''-3ty'+7y=0$$ $$
( t^2 \mathcal D^2 - 3t\mathcal D + 7)y(t) = 0$$ $$
(D(D-1) -3D + 7)y(x) =0$$ $$
(D^2 - 4D + 7)y(x) = 0$$ $$
y'' - 4y' + 7y = 0$$where this is for $y$ in terms of $x$. Now you solve this constant coefficient equation in terms of $x$ and get the solution to the Euler equation by substituting $x=\ln t$.

 

[s3a]

I'm still having some trouble with this.

Before I get back to the notation that I find challenging, I was hoping that I could do this problem in a more familiar way, so could you please point out what I did wrong in my DifferentNotation.pdf file (which is my latest attachment, which is attached in this post)?