# Homework Help: Frobenius Method

1. Feb 24, 2014

### alphamu

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

Use the method of Frobenius, constructing a power series about x = 0,
to find the general solution of this equation (retain terms up to and including those in
square brackets):

4xy'' + 2y' + y = 0 [x7/2]

Note: the solution can be written in closed form, can you see how?

2. Relevant equations

3. The attempt at a solution

set y = $\sum$ anxλ+n to obtain:

4x∑(between n=0 and ∞) an(λ+n)(λ+n-1)xλ+n-2 + 2∑(between n=0 and ∞) an(λ+n)xλ+n-1 + ∑(between n=0 and ∞) anxλ+n = 0

Now relabelling the indices using m = n-1 for the first term, m=n-1 for the second term and m=n for the third term here, in order to get a common coefficient of xλ+m for all terms, will give:

4∑(between m=-1 and ∞) am+1(λ+m+1)(λ+m)xλ+m + 2∑(between m =-1 and ∞) am+1(λ+m+1)xλ+m + ∑(between m=0 and ∞) amxλ+m = 0

Now look at terms when m=-1 (coefficient xλ-1):

4[a02-λ)] + 2[a0λ] = 0

so 4a0λ2 - 4a0λ +2a0λ = 4a0λ2 - 2a0λ = 2a0(λ(λ-1)) = 0

with a0 $\neq$ 0, we have our indicial equation:

λ(λ-1) = 0 so λ1 = 1 and λ2 = 0

The remaining terms give:

∑(between m=0 and ∞) { am+1[(λ+m+1)(λ+m) + (λ+m+1)] + am } xλ+m = 0

so this means ∑(between m=0 and ∞) { am+1[(λ+m+1)2] + am } xλ+m = 0

By taking the am term to the other side and diving by (λ+m+1)2 gives me the recurrence relation:

am+1 = -am / (λ+m+1)2, m= 0, 1, 2, ...

Now looking at λ=1:

am+1 = -am / (m+2)2

and by substituting in values of m=0,1,2,3,4,5,6,7,8 will give me a0, ... , a8 which are the following:

a1 = -a0 / 4, a2 = -a1 / 9, a3 = -a2 / 16 and so on with the denominator being a squared value of (m+2) each time up until a8 = -a7 / 81

Here is the part I'm not sure of, I tried to use these to obtain the first part of the general solution (y1)... by substituting into the general form y = a0 + a1x + a2x2 + ....

I got, y1 = -4a1 - 9a2x - 16a3x2 - 25a4x3 - 36a5x4 - 49a6x5 - 64a7x6 - 81a8x7

and then taking square root of this, i.e. (y1)1/2, gave me the first part of the general solution in a complex form.

Is what I have done so far correct to here? I am confident it is up to the recurrence relation where I'm struggling to form a general solution after this.

I know that you need to use the other value of λ=0 to find the other part of the general solution y2 and your final general solution will be y(x) = y1 + y2.

Again here I am not sure how to find y2 because I've used the recurrence relation to get a0, ... , a8 again but by the Frobenius method if you have two values of λ that differ by an integer (which is the case here), you examine the recurrence relation for λ2 which I have done, but because the recurrence relation is singular at m=1 when λ=0, you seek a second solution y2 by using:

y2 = y1 ln|x| + ∑bnxλ+n

Would this be the correct way to find the second solution and are my working for the first solution correct?

2. Feb 24, 2014

### vela

Staff Emeritus
I got a different indicial equation: $4\lambda^2-2\lambda=0$.