# I need help in a problem

1. May 20, 2008

### momen salah

hi guys it's me you helped me last time in my bonus problem(thank's for that) i need help again naw please it's a hard problem for me :

solve using power series:
(x^2)(y")+y=0

after solving it i stoped at :

an[n^2-n+1]=0

2. May 20, 2008

### exk

If you are trying to solve this equation around the point $x_0=0$ you will run into a problem since that is a singular point of that equation.

3. May 20, 2008

### Crosson

Are you using the method of Frobenius?

Here is the general solution in terms of elementary functions:
$$\sqrt{x} \cos \left(\frac{1}{2} \sqrt{3} \log (x)\right) C_1+\sqrt{x} \sin \left(\frac{1}{2} \sqrt{3} \log (x)\right) C_2$$

4. May 21, 2008

### momen salah

i don't know we only learned how to solve de's when there are no singularities

5. May 21, 2008

### exk

6. May 21, 2008

### momen salah

how can i use this Frobenius method

7. May 24, 2008

### uart

It's quite an interesting problem, I just took the time to solve it now. Normally with a series solution you expect to find a recursion relation for the coefficients, but this one is quite different. Here, as I'm sure you've already discovered (and I'm guessing where you got stuck), there is no recursion relation and at first it may appear that all the coefficients are forced to zero.

If you apply the Frobenius method you should end up with something in the form of,

$$P(r+k) \, \, a_k = 0$$, for each k, where P(.) is a polynomial.

No recursion relation, and all $$a_k$$ must be zero except for at most a finite number corresponding to the zeros of P(.).

If you got this far and it all seemed wrong then don't depair, you're on the right track. Here are some hints to finish it off.

1. Let $$r + k = \omega$$ (or whatever) and solve $$P(\omega) = 0$$

2. If you get complex roots then keep going regardless. (Don't worry as later in the solution you can still restrict the remaining coefficients to force y to be a real function).

3. Don't forget that the complex exponential $$x^{i \beta}$$ can be rewritten as $$e^{i \beta \log(x)}$$.

Follows those hints and you'll get Crosson's solution with surprisingly little effort.

Last edited: May 24, 2008
8. May 24, 2008

### Big-T

Why not "guess" a solution of the form x^r, and end up with solutions $$x^{1/2 \pm i\sqrt{3}/2}$$, which are essentially the same as Crosson ended up with?

9. May 24, 2008

### uart

There's nothing wrong with using an inspired guess (and verify) to solve a DE, but in this instance the OP did explicitly say that he was asked to solve it using a power series method.

10. May 24, 2008

### Big-T

Alright, thanks!