Power Series to solve Second order Differential Equations

In summary, the student is struggling with solving a D.E. with power series and is having troubles with indexing and taking out the first term or two.
  • #1
xWaffle
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Homework Statement



When solving a D.E. with power series, I've encountered something along the lines of:

[itex](2 - r)^{2}g'' = -2[/itex]

Homework Equations



Power Series

The Attempt at a Solution



I know I am just supposed to assume such a series exists, and work from there. But I'm really getting fudged up when it comes to factoring the r's back into the series, which makes the r's in the summation 2 powers higher than just 'n'.

I've reached the end where I have a recursive definition of an+1 in terms of an, but there are ridiculous fractions on the coefficients that I don't think I can generalise for values of n. They don't seem to have a pattern, and that's..a problem.

Is there an easier way to solve this, leaving the r-polynomial as it is? I expanded it and applied g'' to each term, then subbed in the second derivative of the series that represents g(x) [the standard assumed series, anrn]
 
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  • #2
This is just g = g(r), right?
Do you have to use power series? Looks more straightforward to get it into the form g'' = ... and integrate twice.
If you wish to persevere with power series, please post your working.
 
  • #3
It's an assignment, I'm much more than capable of solving this with more traditional means, but I can't submit that work.

I don't have my notebook at the moment but from memory, when using power series, you assume

[itex]y = \sum_{n=0}^{\infty} a_{n}x^{n}[/itex]

Which means

[itex]y' = \sum_{n=1}^{\infty} na_{n}x^{n-1}[/itex]
[itex]y'' = \sum_{n=2}^{\infty} n(n-1)a_{n}x^{n-2}[/itex]

And particularly for the second derivative, re-write the sum so the power of x is the same as for y

[itex]y'' = \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}[/itex]

Now, after putting the sum forms of y and its derivatives back into the equation, I think I fiddled around with it by expanding whatever quantity was multiplied with y'', and replaced all occurrences of y and its derivatives with the appropriate sums.

I guess you could re-write the r's as x's since I wrote x's in the sums instead of r's, so you'd need to factor that variable into the sums when appropriate, which messes with the powers of x again, even though I just fixed them before.

This also leads to problematic situations where different sums are -beginning- at different indices of n. I know I'm supposed to pop out the first few terms til the sum starts at the same index as the others, but I'm unsure about how to do this procedure.

I know the end goal is to get all sums starting at the same index, and all x's in the sums to have the same power, so I can factor out the x's and have one large sum of coefficients.

My problem is, I don't exactly know how to shift indices correctly (because of the a's within the sums) and I don't really know how to take out the first term or two to get the starting index to match with the other sums. I also don't know what I'm supposed to actually do with those coefficients once I get them out of their mother sums.
 

1. What is a power series?

A power series is an infinite series of the form ∑n=0∞ an(x-c)n, where an and c are constants and x is the variable. It is a type of mathematical series used to approximate a function.

2. How are power series used to solve second order differential equations?

A power series can be used to represent a solution to a second order differential equation by substituting it into the equation and solving for the coefficients. This method is known as the power series method and is particularly useful for solving differential equations with non-constant coefficients.

3. What is the radius of convergence for a power series?

The radius of convergence for a power series is the distance from the center of the series (c) to the nearest point where the series converges. It is determined by the ratio test, which states that the series will converge if the limit of |an+1/an| as n approaches infinity is less than 1.

4. Can a power series be used to solve any second order differential equation?

No, a power series can only be used to solve a second order differential equation if the equation can be transformed into a series form. This is not always possible, so other methods may need to be used to solve certain types of differential equations.

5. How accurate are power series solutions to differential equations?

The accuracy of a power series solution depends on the radius of convergence, which in turn depends on the function being approximated. Generally, the more terms included in the series, the more accurate the solution will be. However, power series solutions may only be accurate within a certain range of values for the variable x.

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