Frobenius Method for ODE

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=[tex]\sum[/tex]a_nx^n+r
and plugging it into the ODE, i found

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

i think this can be equated to:
y=(-1/4)a₀e^2x/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?

tiny-tim

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?

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:
[tex]\sum[/tex][tex]^{inf}_{n=0}[/tex](n+r)(n+r-1)a_nx^n+r-2 + [tex]\sum[/tex][tex]^{inf}_{n=0}[/tex](n+r)a_nx^n+r-2 + (1/4)[tex]\sum[/tex][tex]^{inf}_{n=0}[/tex]a_nx^n+r-1 = 0

Simplifying and collecting terms and sum:
=[tex]\sum[/tex][tex]^{inf}_{n=1}[/tex]((n+r)(n+r-1)a_n + (n+r)a_n + (1/4)a_n-1)x^n+r-2 + (0+r)(0+r-1)a₀x^0+r-2 + (0+r)a₀x^0+r-2 = 0

Equating coefficients i found
r² = 0

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

Which gives the recurrence formula:
a_n = [tex]\frac{-a_n-1}{4n²}[/tex] , n=1,2,3,....inf

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

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?

zass

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