Solve the ODE y'' + (3x)/(1+x^2)y' + 1/(1+x^2)y = 0

[Treadstone 71]

I need to determine the fundamental set of solutions of

y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0

in form of a power series centered around x_0=0.

When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?

 

[Physics Monkey]

Try multiplying the whole equation by 1 + x^2.

 

[saltydog]

I need to determine the fundamental set of solutions of
y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0
in form of a power series centered around x_0=0.
When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?

Multiply throughout by (1+x^2) first to get:

(1+x^2)y^{''}+3xy^{'}+y=0

So wouldn't that first term just be:

\sum n(n-1)a_n x^{n-2}+\sum n(n-1)a_nx^n

right?

 

[BerkMath]

There is much to this problem after you get passed your issue, however; I will restrain my response to your question. First off, multiply everything by 1+x^2. Now after subsituting your power series expression for y, y', and y'' you will be left with only 2 powers of x: x^n-2 and x^n. I imagine your question begins here although now in different form. You must make a change of index on the summation containing x^n-2, try m=n-2. After this substitution you will have an expression with a sum with x^m and one with x^n. Since "m" is a dummy index let m=n. It seems circular but it is correct and required. Now you will have an expression in which the only powers of x that appear are x^n. Grouping the coefficients and setting them equal to zero is your next step which I will let you do on your own. It is interesting to note that in order to do this problem you are switching the "shift" of the powers of x to the coefficients an. Not only is this helpful, but it is mandatory when solving the original DE since you must have different an's from which to find your recurrence relations.

 

[Treadstone 71]

If I make a substitution m=n-2, at some point I'll have a term in the form of:

\sum_{n=0}(n+2)(n+1)a_{n+2}x^{n+2}

Are you saying that at this point I can reset the index of x to n and leave a_n the way it is?

 

[BerkMath]

That is not what you will have. The coefficients look correct but its not x^(n+2) it is x^n. Recall: you had x^(n-2), you let m=n-2, therefore
x^((m+2)-2)=x^m with the coefficients you have. Then subsititute back and let m=n.

 

[Treadstone 71]

Ok, got it, I think:

\sum_{n=0}^{\infty}x^n[a_{n+2}(n+2)(n+1)+a_n[n(n-1)+3n+1]]

This implies that

a_{n+2}(n+2)(n+1)+a_n(n+1)^2=0

 

[BerkMath]

yes, now find your recurence relation and your done.