What is the Recursion Relation for Series Solutions in Differential Equations?

[Jamin2112]

Homework Statement

I just have a problem with series solutions when I get to the point of needing to find the recursion relation...and a few other problems.

Homework Equations

Assume y can be written ∑a_nxⁿ

The Attempt at a Solution

So, on y'' + xy' + 2y = 0, I got to the point of

a_n+2=-a_n/(n+1). I wrote out a few n's but still can't figure out exactly what the relation is. A little help, maybe?

On another problem, (1+x2)y'' - 4xy' + 6y = 0, I got the point of

2a₂ + 6a₃x - 4a₁x + 6a₀ + 6a₁x + [n=2 to ∞]∑( (n+1)(n+2)a_n+2 + (n-1)na_n - 4na_n + 6an)xⁿ = 0.

Now what? Detailed help would be much appreciated.

 

[tiny-tim]

Hi Jamin2112!

Assume y can be written ∑a_nxⁿ

The Attempt at a Solution

So, on y'' + xy' + 2y = 0, I got to the point of

a_n+2=-a_n/(n+1). I wrote out a few n's but still can't figure out exactly what the relation is. A little help, maybe?

Hint: 7*5*3*1 = 7*6*5*4*3*2*1/(2³*3*2*1)

On another problem, (1+x2)y'' - 4xy' + 6y = 0, I got the point of

2a₂ + 6a₃x - 4a₁x + 6a₀ + 6a₁x + [n=2 to ∞]∑( (n+1)(n+2)a_n+2 + (n-1)na_n - 4na_n + 6an)xⁿ = 0.

General method is to have one ∑_n=1(…)xⁿ or ∑_n=2(…)xⁿ (whichever seems to work), from which you get a general recursion relation for a_n for n ≥ 1 (or n ≥ 2 or whichever), together with some stray terms that don't fit inside the ∑, and which give you the initial terms.

 

[Jamin2112]

Hi Jamin2112!


Hint: 7*5*3*1 = 7*6*5*4*3*2*1/(2³*3*2*1)


General method is to have one ∑_n=1(…)xⁿ or ∑_n=2(…)xⁿ (whichever seems to work), from which you get a general recursion relation for a_n for n ≥ 1 (or n ≥ 2 or whichever), together with some stray terms that don't fit inside the ∑, and which give you the initial terms.

So the stray terms are the initial conditions. They get set to zero, along with everything inside the sigma, right?

 

[tiny-tim]

Yup!