# Frobenius Method: Solving for General Solution of 4xy'' + 2y' + y = 0 [x7/2]

• alphamu
In summary, the conversation discusses finding the general solution to a differential equation using the Frobenius method. The process involves setting y equal to a power series, relabeling indices, and solving for two values of lambda. The solution can be found by using the recurrence relation and examining the singular point at m=1.
alphamu

## 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 [x7/2]

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

## 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?

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

## 1. What is the Frobenius method?

The Frobenius method is a technique for finding a general solution to a second-order linear differential equation in the form of a power series. This method is particularly useful for equations with singular points, such as when the coefficients are non-analytic functions.

## 2. How does the Frobenius method work?

The Frobenius method involves assuming a solution in the form of a power series, substituting it into the differential equation, and solving for the coefficients. This produces a recurrence relation that can be used to find the coefficients and determine a general solution.

## 3. What is the purpose of the constant term in the power series for the Frobenius method?

The constant term in the power series represents the initial conditions of the differential equation and must be included in order to find a specific solution. Without this term, the solution would not satisfy the given initial conditions.

## 4. What type of differential equations can be solved using the Frobenius method?

The Frobenius method is most commonly used for second-order linear differential equations with variable coefficients. It can also be applied to some higher-order equations, as well as equations with non-constant coefficients.

## 5. How is the solution to a differential equation using the Frobenius method verified?

The solution obtained through the Frobenius method can be verified by substituting it back into the original differential equation and checking that it satisfies the equation. Additionally, the solution can be checked by comparing it to known solutions or using a computer program to plot the solution and confirm its accuracy.

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