# Frobenius Method for ODE

1. Sep 23, 2009

### zass

1. The problem statement, all variables and given/known data

Use the Frobenius method, for an expansion about x=0, to find ONE solution of

xy''+y'+(1/4)y=0

2. Relevant equations

3. The attempt at a solution

starting with an assumption of y1=$$\sum$$anxn+r
and plugging it into the ODE, i found

y=$$\frac{-1}{4}$$$$\sum$$a0xn/(n!)2

i think this can be equated to:
y=(-1/4)a0e2x/x, but when i plug it into the ODE it doesn't equal 0, and thus isn't a solution of the ODE, can anyone see where i've gone wrong?

2. Sep 24, 2009

### tiny-tim

Welcome to PF!

Hi zass! Welcome to PF!
Difficult to say without seeing your full calculations, but did you take into account that the equations for n = 0 and 1 are usually different from the equations for general n?

3. Sep 25, 2009

### zass

I think i accounted for it
Here's my working out, maybe u can see if i've made a mistake:
Using the initial assumption i've found u' and u'', then subbed them into the ode:
$$\sum$$$$^{inf}_{n=0}$$(n+r)(n+r-1)anxn+r-2 + $$\sum$$$$^{inf}_{n=0}$$(n+r)anxn+r-2 + (1/4)$$\sum$$$$^{inf}_{n=0}$$anxn+r-1 = 0

Simplifying and collecting terms and sum:
=$$\sum$$$$^{inf}_{n=1}$$((n+r)(n+r-1)an + (n+r)an + (1/4)an-1)xn+r-2 + (0+r)(0+r-1)a0x0+r-2 + (0+r)a0x0+r-2 = 0

Equating coefficients i found
r2 = 0

And thus the series now becomes
n(n-1)an + nan + (1/4)an-1 = 0

Which gives the recurrence formula:
an = $$\frac{-an-1}{4n2}$$ , n=1,2,3,....inf

then taking the first few terms i found
an = a0/(n!)2
and then just subbed back into the initial assumption made. I can't see anything wrong with what i've done

4. Sep 25, 2009

### tiny-tim

Hi zass!

(have a sigma: ∑ and an infinity: ∞ and if you're using tex, ∞ is \infty, and sub and sup are _{} and ^{} )

You've left out (-1/4)n

does that make it ok?

5. Sep 26, 2009

### zass

ah of course!! for some reason i thought that'd just be a constant.
finally got a solution that fits