# Use the method of Frobenius to

• s3a
In summary, the conversation discusses using the method of Frobenius to find a solution near x=0 of a given equation. The book's solution is using the "dot dot dot" notation, while the speaker prefers the summation notation. The roots of the indicial equation can be different depending on the method used, with the book's solution getting roots of λ_1 = 1 and λ_2 = -2, and the speaker's solution getting roots of r_1 = 0 and r_2 = 1. The correct answer according to the book is provided, while the speaker's answer is also given for comparison. The speaker is unsure if their answer is wrong or if they just got different roots.
s3a

## Homework Statement

Use the method of Frobenius to find one solution near x = 0 of
x^2 y'' + (x^2 + 2x)y' - 2y = 0.

## Homework Equations

Summation notation.

(The solution in my book is using the "dot dot dot" notation which I personally hate a lot since it requires writing a finite amount of infinite terms whereas the summation notation accurately represents all the infinite terms.)

## The Attempt at a Solution

My attempt is attached. Sorry for the occasional bad handwriting, it's because I forget I am writing for you guys. I still think the ugly parts are legible though but I don't know if that's just me.

Can the roots of the indicial equation be different depending on the method used to solve the problem?

Assuming that they can't, the book gets roots λ_1 = 1 and λ_2 = -2 whereas I get roots r_1 = 0 and r_2 = 1.

The correct answer according to the book is:
y_1(x) = a_0 * x * Σn=0,inf (-1)^n 3! x^n /(n+3)!
y_1(x) = 3a_0/x^2 * [2 - 2x + x^2 - 2e^(-x)]

#### Attachments

• MyWork.jpg
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My answer is:y_1(x) = a_0 * x * Σn=0,inf (-1)^n (n+2)! x^n / (2n+3)!y_1(x) = -a_0/x * [1 + x - x^2 + 2e^(-x)]Are my answers wrong or did I just get different roots? Or maybe I am confused and the book's answer is the same as mine?Thanks in advance.

## 1. What is the method of Frobenius?

The method of Frobenius is a mathematical technique used to find a series solution to differential equations with variable coefficients. It involves assuming a solution of the form y(x) = xrn=0 anxn, where r is a constant and an are coefficients to be determined.

## 2. When is the method of Frobenius used?

The method of Frobenius is used when solving differential equations with variable coefficients, especially those that cannot be solved using standard techniques such as separation of variables or substitution.

## 3. What are the steps involved in using the method of Frobenius?

The steps involved in using the method of Frobenius are as follows:
1. Assume a solution of the form y(x) = xrn=0 anxn.
2. Substitute this solution into the differential equation and equate coefficients of like powers of x.
3. Use the recurrence relation to find an+1 in terms of an.
4. Continue this process until the desired number of coefficients have been determined.
5. Plug the determined coefficients back into the assumed solution to get the final series solution.

## 4. What are some limitations of the method of Frobenius?

The method of Frobenius may not always provide a solution, as it relies on guessing an appropriate form for the solution. It also may not provide a valid solution if there are repeated roots or complex roots involved. Additionally, the series solution may not converge for all values of x, limiting the applicability of this method.

## 5. Can the method of Frobenius be used for any type of differential equation?

No, the method of Frobenius can only be used for certain types of differential equations, specifically those with variable coefficients. It may not be applicable for equations with constant coefficients or non-linear equations.

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