(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(1+x^{2})y'' - 4xy' + 6y = 0

2. Relevant equations

I'm going to assume y can be written as [n=0 to ∞] ∑a_{n}x^{n}

3. The attempt at a solution

y = [n=0 to ∞] ∑a_{n}x^{n}

----> y' = [n=0 to ∞] ∑(n+1)a_{n+1}x^{n}

----> y'' = [n=0 to ∞] ∑(n+2)(n+1)a_{n+2}x^{n}

----> (1+x^{2})∑(n+2)(n+1)a_{n+2}x^{n}-4x∑(n+1)a_{n+1}x^{n}+ 6∑(n+1)a_{n+1}x^{n}= 0.

My problem is that, after multiplying out the functions in front of y,y',y'', everything is summed from n=0 to ∞ but the the exponents on the x's don't match up. For example, one term looks like ∑(n+1)(n+2)a^{n+2}x^{n+2}. I can't just do a transformation j=n+2, making it

[j=-2 to ∞] ∑(j-1)ja_{j}x^{j}= (-3)(-2)a_{-2}x^{-2}+ (-2)(-1)a_{-1}x^{-1}+ 0 + 0 + (1)(2)a_{2}x^{2}+ (2)(3)a_{3}x^{3}+ ...

because then I have those darned a_{-2},a_{-1}that screw everything up. In other words, I want everything summed from n=0 to ∞andonly x^{n}being summed. Then I can figure out the recurrence relation, the radius of convergence, etc. Suggestions are appreciated.

NEXT QUESTION:

On another problem, I was able to get to the point of

[n=0 to ∞] ∑x^{n}{(n+2)(n+1)a_{n+2 }+ na_{n}+ 2a_{n}}=0.

For this equation to be satisfied for all x, the coefficient of each power of x must be zero; hence I conclude that

(n+2)(n+1)a_{n+2 }+ na_{n}+ 2a_{n}= 0.

I'm now trying to figure out the recurrence relationship.

I write out

2a_{0}+ 2a_{2}= 0 -----> a_{2}=-a_{0}

3a_{1}+ 6a_{3}= 0 ------> a_{3}=(-1/2)a_{1}

4a_{2}+ 6a_{3}= 0 ------> a_{4}=-(1/3)a_{2}=(1/3)a_{0}

5a_{3}+ 2a_{5}= 0 ------> a_{5}=-(1/4)a_{3}=(1/8)a_{1}

6a_{4}+ 30a_{6}= 0 -----> a_{6}=-(1/5)a_{4}=(-1/15)a_{0}

7a_{5}+ 42a_{7}= 0 ------>a_{7}=(-1/6)a_{5}=(-1/48)a_{1}

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---------> a_{n}= a_{2k}= (-1)^{k}* a_{0}/((2k-1)(2k-3)) ?????????

I dunno. Does that seem right? I'm just trying to figure it out in my head. Supposing this is correct, what then do I do with the a_{2k+1}stuff--you know, the a_{1},a_{3},a_{5},...

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# Homework Help: Series Solutions of Second Order Linear Equations

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