# A harder Frobenius Differential

• Mattofix
In summary, the conversation was about using the Frobenius method to solve the equation xu'' + 2y' + xy = 0. The poster had solved this method before but found this problem more challenging. They shared their attempt at a solution, which included using summation notation and trying to remove n=0 and n=1 to find the indicial polynomial. However, they were unsure if their solution was correct and asked for clarification. In the end, they were able to solve the problem.
Mattofix

## Homework Statement

xu'' + 2y' + xy = 0

## Homework Equations

http://en.wikipedia.org/wiki/Frobenius_method

## The Attempt at a Solution

Ok, so i have managed the Frobenius method in the past, but this seems harder...

$$\sum_{n=0}^{\infty} a_n(n+c)(n+c-1)x^{\ n+c-2} + 2\sum_{n=0}^{\infty} a_n(n+c)x^{\ n+c-2} + \sum_{n=0}^{\infty} a_nx^{\ n+c} = 0$$

$$\sum_{n=0}^{\infty} a_n(n+c)(n+c-1)x^{\ n+c-2} + 2\sum_{n=0}^{\infty} a_n(n+c)x^{\ n+c-2} + \sum_{n=2}^{\infty} a_{\ n-2}x^{\ n+c-2} = 0$$

i cannot simply remove n=0 as i would if there were only a difference of one between the sumation limits, here i would have to remove n=0 and n=1 which means that i would not have a quadratic multiplied by $$a_0$$to solve, to find the c values. I would have one quadratic multiplied $$a_0$$ by and one by $$a_1$$...

I meant xy'' + 2y' + xy = 0...

To clarify, from the wikipedia article and looking at the example,

'We can take one element out of the sums that start with k=0 to obtain the sums starting at the same index.'

'We obtain one linearly independent solution by solving the indicial polynomial r(r-1)-r+1 = r2-2r+1 =0 which gives a double root of 1'

I have two equations if i remove n=0 and n=1, what shall i do to get my indicial polynomial?

Here is the question:

and my solution:

am i right? is the question wrong? please can someone clarify this.

... :(

f uckyall bitchez - i solved it.

## 1. What is a harder Frobenius differential?

A harder Frobenius differential is a type of differential equation that is solved using the Frobenius method. It involves a power series solution with a specific form, including terms that are multiplied by the derivative of the variable. This method is often used for solving second-order linear differential equations with variable coefficients.

## 2. How is a harder Frobenius differential different from a regular Frobenius differential?

A regular Frobenius differential involves solving a differential equation with a specific form of power series solution, while a harder Frobenius differential adds terms that are multiplied by the derivative of the variable. This extra term makes the equation more difficult to solve, hence the name "harder" Frobenius differential.

## 3. What types of problems can be solved using a harder Frobenius differential?

A harder Frobenius differential is typically used to solve second-order linear differential equations with variable coefficients. These types of equations can arise in many areas of science and engineering, such as physics, chemistry, and economics.

## 4. What is the process for solving a harder Frobenius differential?

The process for solving a harder Frobenius differential involves first identifying the equation as a second-order linear differential equation with variable coefficients. Then, the power series solution is substituted into the equation, and the coefficients are determined by comparing terms and equating them to zero. This process may require several iterations to find the correct coefficients.

## 5. Are there any limitations to using a harder Frobenius differential for solving equations?

While the harder Frobenius differential method is powerful and can solve many types of equations, it does have limitations. This method is most effective for equations with variable coefficients that can be expressed in terms of a power series solution. Other types of equations, such as those with non-polynomial solutions, may require different methods for solving.

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