# Power Series Diff. Eq.

1. Apr 11, 2006

### kahless2005

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?

2. Apr 12, 2006

### dextercioby

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

Daniel.

3. Apr 12, 2006

### HallsofIvy

Staff Emeritus
In other words:
Can you change
$$\Sigma_{n=0}^\infty C_n x^{n+1}$$
to
$$\Sigma_{n=1}^\infty C_{n-1}x^n$$

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
$$\Sigma_{n=0}^\infty C_n x^{n+1}$$
is the same as
$$\Sigma_{j=1}^\infty C_{j-1} x^j$$
and since j is just a dummy index, you can certainly replace it with n.