Find the power series?

Question

Find the power series in x for the general solution of (1+x^2)y"+6xy'+6y=0.

 

Please show all your work step by step.

Answer 1

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Answer 2

Yang,

 

Do you know how to get started? I sent a response to an earlier question similar to this one, and you never replied back.

 

Assume the solution has the form of a power series centered at 0:

 

y(x) = ∑_n=0^∞ a_nxⁿ

 

Take the first two derivatives w.r.t x term-by-term:

 

y′(x) = ∑_n=1^∞ na_nx^n-1  and  y″(x) = ∑_n=2^∞ n(n-1)a_nx^n-2.

 

If you have any questions about why the derivatives have index starting at 1 and 2, you should ask.

 

Plug these into the DE and simplify.

 

(1 + x²)y″ = ∑_n=2^∞ n(n-1)a_nx^n-2 + x² ∑_n=2^∞ n(n-1)a_nx^n-2 = ∑_n=2^∞ n(n-1)a_nx^n-2 + ∑_n=2^∞ n(n-1)a_nxⁿ

 

   = ∑_k=0^∞ (k+2)(k+1)a_k+2x^k + ∑_n=2^∞ n(n-1)a_nxⁿ

 

In the last step, I reindexed the first sigma so it would have exponent n instead of n-2. I did this by letting k = n-2, so that n = k+2, started the new sum at 0, and rewrote the sigma using k rather than n.

The next step (before I send this to you to think about) is to write these two sigmas using one sigma. 

 

(1 + x²)y″ = 2a₂ + 6a₃x + ∑_k=2^∞ [(k+2)(k+1)a_k+2 + k(k-1)a_k]x^k.

 

What I did in the last step was to write the first two terms in the first sigma (k=0 and k=2) separately, because the other terms can be combined with the second sigma which starts with the term x². I then combined the remaining terms into a single sigma.

 

I will not post any additional help on this problem until you send a reply letting me know you are receiving it or you have any questions about what I sent here or in my last reply.

 

Sorry, but this takes a lot of patient effort to send in WyzAnt and I would be better off helping some other student who will give me feedback on whether my hints are being helpful. If you prefer not to have me send you any more help on your DE questions, please let me know this. So far, I appear to be the only tutor who has responded to your posts and I do not want to block out someone else who may be of better help to you.

 

Thanks for your feedback and I look forward to hearing from you soon. Have a happy fourth of July holiday (if you celebrate it)!