# Help with frobenius

1. May 8, 2005

### jason17349

My problem: find the first solution and use it to find the second solution for

x^2*y"-x*y'+(x^2+1)y=0

assuming y=summation from n=0 to infinity for An*x^n+r

substituting and solving gives me r=1 and a general equation: An=A(n-2)/((n+r)*(n+r-2)+1) for n >= 2

plugging r into my general equation gives An=A(n-2)/((n+1)*(n-1)+1) for n >= 2

plugging n into this I get y=A0*x+(1/4)A0*x^3+(1/64)A0*x^5+(1/2304)A0*x^7........ this is y1

now y2=y1*v

I'm not entirely sure what to do after this because I'm unable to reduce y1 to a simple summation which is the only way I've seen this problem done before

Last edited: May 8, 2005
2. May 8, 2005

### inha

Maybe you could try to write A_n in product form to see if something more manageable pops out with a nice selection* of A_0.

*You know, the kind of selections people who already know the answer to the problem always do just to mock us mere mortals.

3. May 8, 2005

### jason17349

Using the reduction formula I came up with this solution:

y2=y1*integral(x*(y1)^(-2)*dx)

I don't know how to write math symbols in here so I attached a picture that is easier to understand.

does this seem like the correct solution? I'm also concerned about my answer for y1. The index value and power particularly.

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4. May 8, 2005

### Hurkyl

Staff Emeritus