How to Solve Nonhomogeneous Euler Equations with Absolute Values?

[S.N.]

Homework Statement

Solve the IVP

(x^2)y'' + 4xy' - 40y = x^6

for y(1) = 10, y'(1) = 1

Homework Equations

not so much "equations" but here I try to use variation of parameters to get the particular solution.

The Attempt at a Solution

FOR THE HOMOGENEOUS SOLUTION:
using the substitution y = x^r I get a characteristic equation of

r^2 - 3r -40 = 0
so,
(r-8)(r+5) = 0

and then, the homogeneous solution will be

y_h = c₁|x|^8 + c₂|x|^-5

FOR THE PARTICULAR SOLUTION:

Here is where I run into trouble. I'm unsure of how this part is usually done, but I'm guessing that since the original problem does not have constant coefficients, we are ofrced to use variation of parameters (can someone confirm this? It seems right to me but maybe there's a trick I'm missing here).

so, first I get the wronskian, which is the determinant of this 'matrix':

|x|⁸ |x|^-5
8|x|⁷ -5|x|^-6

I get -5|x|² - 8|x|² = -13|x|²

then here's my main problem (or maybe my problem is earlier -- not sure if I'm doing this right). I get to the variation of parameters part, and there's two integrals I have to do:

integral of |x|⁸x⁶ divided by 13|x|²

and integral of |x|^-5x⁶ divided by 13|x|²

here is where I get confused: how do I deal with the integrals with absolute values? Did I do something completely wrong or is there something I'm not seeing? Is there some way to avoid this mess? Thanks for any help, my textbook never gets into the specifics of how to deal with an euler equation when it isn't homogeneous

 

[hunt_mat]

Why not try the particular solution of:
<br /> y_{p}=a_{1}x^{6}+a_{2}x^{5}+a_{3}x^{4}+a_{4}x^{3}+a_{5}x^{2}+a_{6}x+a_{7}<br />

 

[S.N.]

2 things:

1) I looked at this intuitively and just thought "well, why not just guess Ax^6 because I'll just have 3 things multiplied by x^6 on the LHS afterward"? And it worked out. It makes perfect sense.

2) To hunt_mat: now that I read your post, is this what you're getting at?

 

[LCKurtz]

2 things:

1) I looked at this intuitively and just thought "well, why not just guess Ax^6 because I'll just have 3 things multiplied by x^6 on the LHS afterward"? And it worked out. It makes perfect sense.

That is exactly what to do.

 

[hunt_mat]

2 things:

1) I looked at this intuitively and just thought "well, why not just guess Ax^6 because I'll just have 3 things multiplied by x^6 on the LHS afterward"? And it worked out. It makes perfect sense.

2) To hunt_mat: now that I read your post, is this what you're getting at?

It was exactly what I was getting at.