- #1
Jamin2112
- 986
- 12
Homework Statement
(1+x2)y'' - 4xy' + 6y = 0
Homework Equations
I'm going to assume y can be written as [n=0 to ∞] ∑anxn
The Attempt at a Solution
y = [n=0 to ∞] ∑anxn
----> y' = [n=0 to ∞] ∑(n+1)an+1xn
----> y'' = [n=0 to ∞] ∑(n+2)(n+1)an+2xn
----> (1+x2)∑(n+2)(n+1)an+2xn -4x∑(n+1)an+1xn + 6∑(n+1)an+1xn = 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)an+2xn+2. I can't just do a transformation j=n+2, making it
[j=-2 to ∞] ∑(j-1)jajxj = (-3)(-2)a-2x-2 + (-2)(-1)a-1x-1 + 0 + 0 + (1)(2)a2x2 + (2)(3)a3x3 + ...
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 ∞ and only xn 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 ∞] ∑xn{(n+2)(n+1)an+2 + nan + 2an}=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)an+2 + nan + 2an = 0.
I'm now trying to figure out the recurrence relationship.
I write out
2a0 + 2a2 = 0 -----> a2=-a0
3a1 + 6a3 = 0 ------> a3=(-1/2)a1
4a2 + 6a3 = 0 ------> a4=-(1/3)a2=(1/3)a0
5a3 + 2a5 = 0 ------> a5=-(1/4)a3=(1/8)a1
6a4 + 30a6 = 0 -----> a6=-(1/5)a4=(-1/15)a0
7a5 + 42a7 = 0 ------>a7=(-1/6)a5=(-1/48)a1
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---------> an = a2k = (-1)k * a0/((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 a2k+1 stuff--you know, the a1,a3,a5,...