Consider a power series solution to differential equation y′′+(1 +x^2)y′+(1−x^2)y= 0

[promitheus]

Hello, having a bit of trouble with how to proceed with this power series problem.

Here is the question:

Consider a power series solution to differential equation y′′+(1 +x^2)y′+(1−x^2)y= 0.
(a) Derive a recurrence relation for the coefficients an.
(b) Now suppose a0= 1 and a1= 0. Use your answer to (a) to evaluate a2, a3, a4, and a5

I'm having trouble with getting a useful recurrence relation. I suspect I have made some rookie mistake in my attempt to match the power x for all terms. Any guidance would be much appreciated. I have attached my working.

 

[HallsofIvy]

I don't know that you have made a mistake. Working it out myself I have \(\displaystyle \sum n(n-1)a_nx^{n-2}+ \sum na_na^{n-1}+ \sum a_nx^{n+1}- \sum a_nx^n- \sum a_nx^{n+2}= 0\).

To get the same power of x, so that we combine the exponents, in the first sum, let j= n-2 so that n= j+2 and the first sum becomes \(\displaystyle \sum (j+2)(j+1)a_{j+2}x^j\).

In the second sum, let j= n- 1 so n= j+ 1 and the second sum becomes \(\displaystyle \sum (j+1)a_{j+1}x^j\)

In the third sum, let j= n+1 so that n= j- 1 and the third sum becomes \(\displaystyle \sum (j-1)a_{j-1}x^j\).

In the fourth sum, let j= n so the fourth sum becomes \(\displaystyle \sum a_jx^j\).

In the fifth sum, let j= n+2 so n= j- 2 and the fifth sum becomes \(\displaystyle \sum a_{j-2}x^j\).

Putting those all together, we have \(\displaystyle \sum ((j+2)(j+1)a_{j+2}+ (j+1)a_{j+1}+ (j-1)a_{j-1}+ a_j- a_{j-2})x^j\) and so must have \(\displaystyle (j+2)(j+1)a_{j+2}+ (j+1)a_{j+1}+ (j-1)a_{j-1}+ a_j- a_{j-2}= 0\). We can write that as \(\displaystyle a_{j+2}= \frac{1}{(j+2)(j+1)}\left(a_{j-2}- (j+1)a_{j+1}- (j-1)a_{j-1}- a_j\right)\).

That's your recurrence relation! Since this is a second order equation, there will have to be two additional constants. Typically that will be the values of y and y' at x= 0 so the values of \(\displaystyle a_0\) and \(\displaystyle a_1\). Use the first two equations you give to determine \(\displaystyle a_2\) and \(\displaystyle a_3\) and then, with j= 2, \(\displaystyle a_4= \frac{1}{6}\left(a_0- 3a_3- a_1- a_2\right)\), etc.

 

[promitheus]

Thanks a bunch for the reply. I got confused because I wrote my recurrence relation in terms of a_n. Nice to be reassured I'm on track.

Noted a slight error in your last statement with j=2, a₄ = 1/12 (a₀ - 3a₃ - a₁ - a₂)

Cheers!