Question

Hi! I'm having trouble with an ODE. I can solve for solutions in the form
\(x^2y''+\alpha y' +\beta y=0\)

But now I have
\((x-1)^2y''+8(x-1)y' +12 y=0\)
with the solution
\(y=\dfrac{c_1}{(x-1)^3}+\dfrac{c_2}{(x-1)^4}\)

Answer 1

I considered a substitution, but then (I probably used illegal moves, but) \(\dfrac{d^2y}{du^2}\) where \(u=x-1\) is undefined.

Answer 2

You're a better physicist than me Eric. Can't help here.

Answer 3

power series method

Answer 4

Haha, thanks. It's hard to compare physics knowledge! Anyway, this is mathematics!

Answer 5

Thank you @Loser66 , I'll look into that!

Answer 6

what's wrong with u = x - 1?

Answer 7

For series solutions, don't they need to be around a point? Can I just assume the point where \(x=0\) has a radius of convergence of \(\infty\)? The problem indicates that the solution doesn't need to work at the singular points.

Answer 8

nope, regular singular point is x =1 because it makes y"=0

Answer 9

If I use \(u=x-1\), then I can't have \(\dfrac{\text dy}{\text dx}\) and still solve it with the shortcut

\(y=c_1x^{r_1}+c_2x^{r_2}\)
where
\(r_1\) and \(r_2\) are the roots of
\(r^2+(\alpha -1) r+\beta=0\)
assuming two distinct real roots?

Answer 10

The singular point is \(x=1\), yep. But I just need to find a solution for all \(x\neq1\).

Answer 11

So, I guess I can try the power series solution about \(x=0\) and suppose that it would converge for all values that aren't \(x=1\)?

Answer 12

\(\dfrac{\text dy}{\text du}=\dfrac{\text dy}{\text dx}\dfrac{\text dx}{\text du}\)
where

\(\dfrac{\text du}{\text dx}=\dfrac{\text d}{\text dx}\left(x-1\right)=1\)\(\implies\dfrac{\text dx}{\text du}=1\)
I don't know if that's allowed...

Either way, \(u+1=x\), so \(\dfrac{\text dx}{\text du}=1\)

So then
\(\dfrac{\text dy}{\text du}=\dfrac{\text dy}{\text dx}\dot\ 1=y'\)
That works. But a second derivative would cause \(\dfrac{\text d^2y}{\text du}=0\), I think. I erased what I had.

I'll seek a power series solution, I suppose.

Answer 13

Haha, just now found \((x-1)^2y''\).

The next ones should be quicker.

Answer 14

i think you can handle from this

Answer 15

Thank you :)

Answer 16

the rest is easy, right? hehehe

Answer 17

Yeah! Thank you very much! I'll get to that point on my own and continue from there. When I take the derivative of the series, though, I shift the index to eliminate the term that became zero upon differentiating.

Answer 18

And then I shift indeces so that all \(x\) are to the \(n\) power, but, now that I think about it, that just makes more work.

Answer 19

Or maybe it's necessary. I can't be sure just by looking at it. But I have to much work to do for me to use my time understanding what will and won't work, unfortunately.

Answer 20

My prof said that if you want to make them have the same form, just convert all of the indices in a sum, for example in y", its sum is
\[\sum_{n=0}^{\infty} n(n-1) a_n x^{n-2}\]
if you want all indices = n, then, replace them ALL by +2 into the indices
\[\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n}\]
Dat sit

Answer 21

so that you can factor x^n out, and the leftover is a power series of coefficients

Answer 22

Right. That's what I do. I also do this:

Assume that \(y\) is a power series solution about \(x=0\).
Then
\[y=\sum_{n=0}^\infty a_nx^n\]
and
\[\dfrac{\text dy}{\text dx}=\sum_{n=1}^\infty a_nnx^{n-1}=\sum_{n=0}^\infty a_{n+1}(n+1)x^{n}\]
and
\[\dfrac{\text d^2y}{\text dx^2}=\dfrac{\text dy}{\text dx}\left[\sum_{n=1}^\infty a_nnx^{n-1}\right]=\sum_{n=2}^\infty a_nn(n-1)x^{n-2}=\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)x^n\]

I change the starting index according to the derivatives I've taken. But I don't think that is necessary.

Answer 23

Rather, I just don't know.

Answer 24

I didn't reach a recursive relation... I got that \(a_0\), \(a_1\), and \(a_2\) are all dependent on each other, and \(a_n=0\) for \(n\ge2\). So \(\forall n\), \(a_n=0\). Bleh...

Answer 25

Thank you for your help, everyone! I don't need to finish this, so maybe I'll try to tackle it later!