- #1

alphamu

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## Homework Statement

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 [x

^{7/2}]

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

## Homework Equations

## The Attempt at a Solution

set y = [itex]\sum[/itex] a

_{n}x

^{λ+n}to obtain:

4x∑(between n=0 and ∞) a

_{n}(λ+n)(λ+n-1)x

^{λ+n-2}+ 2∑(between n=0 and ∞) a

_{n}(λ+n)x

^{λ+n-1}+ ∑(between n=0 and ∞) a

_{n}x

^{λ+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 x

^{λ+m}for all terms, will give:

4∑(between m=-1 and ∞) a

_{m+1}(λ+m+1)(λ+m)x

^{λ+m}+ 2∑(between m =-1 and ∞) a

_{m+1}(λ+m+1)x

^{λ+m}+ ∑(between m=0 and ∞) a

_{m}x

^{λ+m}= 0

Now look at terms when m=-1 (coefficient x

^{λ-1}):

4[a

_{0}(λ

^{2}-λ)] + 2[a

_{0}λ] = 0

so 4a

_{0}λ

^{2}- 4a

_{0}λ +2a

_{0}λ = 4a

_{0}λ

^{2}- 2a

_{0}λ = 2a

_{0}(λ(λ-1)) = 0

with a

_{0}[itex]\neq[/itex] 0, we have our indicial equation:

λ(λ-1) = 0 so λ

_{1}= 1 and λ

_{2}= 0

The remaining terms give:

∑(between m=0 and ∞) { a

_{m+1}[(λ+m+1)(λ+m) + (λ+m+1)] + a

_{m}} x

^{λ+m}= 0

so this means ∑(between m=0 and ∞) { a

_{m+1}[(λ+m+1)

^{2}] + a

_{m}} x

^{λ+m}= 0

By taking the a

_{m}term to the other side and diving by (λ+m+1)

^{2}gives me the recurrence relation:

a

_{m+1}= -a

_{m}/ (λ+m+1)

^{2}, m= 0, 1, 2, ...

Now looking at λ=1:

a

_{m+1}= -a

_{m}/ (m+2)

^{2}

and by substituting in values of m=0,1,2,3,4,5,6,7,8 will give me a

_{0}, ... , a

_{8}which are the following:

a

_{1}= -a

_{0}/ 4, a

_{2}= -a

_{1}/ 9, a

_{3}= -a

_{2}/ 16 and so on with the denominator being a squared value of (m+2) each time up until a

_{8}= -a

_{7}/ 81

Here is the part I'm not sure of, I tried to use these to obtain the first part of the general solution (y

_{1})... by substituting into the general form y = a

_{0}+ a

_{1}x + a

_{2}x

^{2}+ ...

I got, y

_{1}= -4a

_{1}- 9a

_{2}x - 16a

_{3}x

^{2}- 25a

_{4}x

^{3}- 36a

_{5}x

^{4}- 49a

_{6}x

^{5}- 64a

_{7}x

^{6}- 81a

_{8}x

^{7}

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 y

_{2}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 λ

_{2}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| + ∑b

_{n}x

^{λ+n}

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

Can anyone help me please?