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Differential Eq. and power series

  • Thread starter Gogeta007
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  • #1
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Homework Statement



y'' + x2y' + xy = 0

Homework Equations



using power series:
y'' = [tex]\Sigma[/tex]cnn(n-1)xn-2 (n = 2 -> infinity)
y' = [tex]\Sigma[/tex]cn(n-1)xn-1 (n = 1 -> infinity)
y = [tex]\Sigma[/tex]cnxn (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]cnn(n-1)xn-2 + [tex]\Sigma[/tex]cn(n-1)xn+1 + [tex]\Sigma[/tex]cnxn+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=4cnn(n-1)xn-2 + [tex]\Sigma[/tex]cn(n-1)xn+1 + c0x + [tex]\Sigma[/tex]n=1cnxn+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 xk { ck+2(k+2)(k+1) + ck-1(k-1) = ck-1


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

===========

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 = -{ck-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=0n-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=0n-1(3m+2) } / 10*9*7*6*4*3


I couldnt find a pattern for the denominators. . . im 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.
 

Answers and Replies

  • #2
vela
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The relation for the coefficients relates ck to ck+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.
 
  • #3
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thanks! and how do you know where to start your n for the sumation? and where did you get k+3 from?
 
  • #4
vela
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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 cn=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 cn=1/(n+1), I would start the summation at n=0.
 

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