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Power Series Diff. Eq.

  • #1
Given:
y'+2xy=0
Find:
Write sereis as an elementary function

My solution so far:
y=[Sum n=0, to infinity]C(sub-n)*x^n
y'=[Sum n=1, to infinity]n*C(sub-n)*x^(n-1)


y' can be transformed into:
=[Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n

([Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n) + 2x([Sum n=0, to infinity]C(sub-n)*x^n)=0

([Sum n=0, to infinity](n+1)*C(sub-n+1)*x^n) +2([Sum n=0, to infinity]C(sub-n)*x^(n+1))=0


My question:
Can I transform [Sum n=0, to infinity]C(sub-n)*x^(n+1) into ([Sum n=1, to infinity]C(sub-n-1)*x^(n)?

If so, then y(x)=([Sum n=,0. to infinity]((-1)^(n+1)*C(sub-o)*x^(n+1)*2^(n-1))/n!

How do I transform that into an elementary function?

Sorry about the ugly typing...
 

Answers and Replies

  • #2
dextercioby
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Why don't you simply integrate the ODE and then write the series expansion for the solution...?

Daniel.
 
  • #3
HallsofIvy
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kathless2005 said:
My question:
Can I transform [Sum n=0, to infinity]C(sub-n)*x^(n+1) into ([Sum n=1, to infinity]C(sub-n-1)*x^(n)?
In other words:
Can you change
[tex]\Sigma_{n=0}^\infty C_n x^{n+1}[/tex]
to
[tex]\Sigma_{n=1}^\infty C_{n-1}x^n[/tex]

I think you will find it easier to use a different letter for the index:
Let j= n+1. Then n= j-1 so Cn becomes Cj and xn+1 becomes xj[/sub]. Of course when n= 0, j= 1 so
[tex]\Sigma_{n=0}^\infty C_n x^{n+1}[/tex]
is the same as
[tex]\Sigma_{j=1}^\infty C_{j-1} x^j[/tex]
and since j is just a dummy index, you can certainly replace it with n.
 

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