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Differential Eq- Power Series Solution

  • Thread starter sami23
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  • #1
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Find a power series sol'n: (x2-1)y'' + 3xy' + xy = 0

Homework Equations


let y = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=0) Cnxn
let y' = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=1) nCnxn-1
let y'' = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)Cnxn-2

The Attempt at a Solution


I wrote the differential eq as: x2y''-y''+3xy'+xy=0

Substituting back into the differential eq and multiplying the x2 gives:
[tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)Cnxn - [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)Cnxn-2 + 3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=1) nCnxn + [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=0) Cnxn+1 = 0

For the first 2 terms I let k=n-2 , n=k+2 which would give:
[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)Ck+2xk+2 - [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)Ck+2xk

For the 3rd and 4th term I let k=n which would give:
3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=1) kCkxk + [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) Ckxk+1

The new series would be:
[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)Ck+2xk+2 - [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)Ck+2xk + 3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=1) kCkxk + [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) Ckxk+1

I don't know how to make the starting value of each index the same (from [tex]\infty[/tex] to k=0) and how would I get to the general solution?
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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First off a tip. Write your summations like this: \sum_{n=a}^\infty. This will look like [itex] \sum_{n=a}^\infty[/itex], which is a lot neater. Secondly put the tex brackets around your entire equation, not just the summation symbols.

This will make your equation look like:
[tex]\sum_{n=0}^\infty c_n x^n[/tex]

Let's start with a simple example. Take [itex]\sum_{n=1}^\infty c_{n-1}x^{n-1}[/itex]. We want to write this series as [itex]\sum_{n=0}^\infty c_k x^k[/itex]. Lets take a look at the first series.

[tex]
\sum_{n=1}^\infty c_{n-1}x^{n-1}=c_0 x^0+c_1 x^1+c_2 x^2+...
[/tex]

We obviously want [itex]\sum_{n=0}^\infty c_k x^k[/itex] to be equal to this. So how would you express k in terms of n?

[tex]\sum_{n=0}^\infty c_k x^k=c_0 x^0+c_1 x^1+c_2 x^2+...[/tex]?
 
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