Differential Eq. and power series

[Gogeta007]

Homework Statement

y'' + x²y' + xy = 0

Homework Equations

using power series:
y'' = [tex]\Sigma[/tex]c_nn(n-1)x^n-2 (n = 2 -> infinity)
y' = [tex]\Sigma[/tex]c_n(n-1)x^n-1 (n = 1 -> infinity)
y = [tex]\Sigma[/tex]c_nxⁿ (n = 0 -> infinty)

The Attempt at a Solution

by setting the above equation with its corresponding sumation formulas I get the following:

[tex]\Sigma[/tex]c_nn(n-1)x^n-2 + [tex]\Sigma[/tex]c_n(n-1)xⁿ⁺¹ + [tex]\Sigma[/tex]c_nxⁿ⁺¹

to make all the x's start at the same power I pull out 2 terms from the first sumation and 1 term from the last eq. so all x's start at the power of 2.
this yields:

2c₂ + 6c₃x + [tex]\Sigma[/tex]_n=4c_nn(n-1)x^n-2 + [tex]\Sigma[/tex]c_n(n-1)xⁿ⁺¹ + c₀x + [tex]\Sigma[/tex]_n=1c_nxⁿ⁺¹

on the first sumation k=n-2, so n= k+2, on the second sumation k=n+1 and n=k-1 and the same for the third sumation.
substituting k's in their respective sumations we get:

c₀x+2c₂+6c₃x + [tex]\Sigma[/tex]_k=2 x^k { c_k+2(k+2)(k+1) + c_k-1(k-1) = c_k-1


from the last two terms we can factor a c_k-1 and cancel out the 1's left leaving us only with kc_k-1

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from the identity that the coefficients should add up to zero, we conclude that c₂=0 and c₃= -c₀/6

for the sumation part, I solve for c_k+2 = -{c_k-1(k)}/(k+2)(k+1)

and plugging in values of k (1,2,3,4,5,6,7,8,9,10) I can see that k=3 = 0, k=6 = 0, k=9 = 0

for the other values I get sumations with c₀ and sumations with c₁
Im writing them the way I saw the teacher writing them (since the book does without the multiplication symbol (big pi))

I get the following:

c₀ [tex]\Sigma[/tex]_n=?? { (-1)ⁿ [tex]\Pi[/tex]_m=0^n-1(3m+1) } / 12*11*9*8*6*5*3*2

where big pi starts at m =0 and ends at m = n-1

for c_0
and for c_1

c₁ [tex]\Sigma[/tex]_n=?? { (-1)ⁿ [tex]\Pi[/tex]_m=0^n-1(3m+2) } / 10*9*7*6*4*3


I couldn't find a pattern for the denominators. . . I am thinking its two patterns ( for even and odds). . .anyways I'll do that later, my question is the following:

How do I know where to start the sumations of sigma in the answer, you know n=something ?? (why?)

in the book it says that the c₀ sumation has the x³,x⁶,x⁹. . . and that c₁ sumation has the x⁴,x⁷,x¹⁰

how do I know which sumation has what powers of x?

Some of the answers (in class) start with 1 + [tex]\Sigma[/tex]. . . or with x + [tex]\Sigma[/tex]. . .
how do I know when to pull out the first term? and again, on what n to start for the sumation?

Thanks a lot for reading trough this and thanks in advance for your help.

 

[vela]

The relation for the coefficients relates c_k to c_k+3, so

c₀ -> c₃ -> c₆ -> c₉ -> ...
c₁ -> c₄ -> c₇ -> c₁₀ -> ...
c₂ -> c₅ -> c₈ -> c₁₁ -> ...

So the x⁰, x³, x⁶, ... terms are all proportional to c₀. Likewise, the x¹, x⁴, x⁷, ... terms are all proportional to c₁.

Unless there's something special about the first few terms, there's no reason to pull them out of the summation.

 

[Gogeta007]

thanks! and how do you know where to start your n for the sumation? and where did you get k+3 from?

 

[vela]

I just substituted k+1 for k in the recurrence relation you derived to get rid of that awkward k-1, k+2 combination.

You start n at whatever value works, the one that will give you the correct first term of the series. For example, consider the harmonic series 1/1+1/2+1/3+1/4+... If I were to write the n-th term as c_n=1/n, I'd start n at 1 because that's the value that would give me 1/1 as the first term. If instead I wrote c_n=1/(n+1), I would start the summation at n=0.