## Frobenius Method for ODE

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?

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Hi zass! Welcome to PF!
 Quote by zass … 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?
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?
 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

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## Frobenius Method for ODE

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