Question

Not understanding the substitution in this DE solution by series problem.

Answer 1

\[a _{0} \sum_{n=0}^{\infty}\frac{ (3x)^{2n} }{ (2n)! } + \frac{ a _{1} }{ 3 }\sum_{n=0}^{\infty}\frac{ (3x)^{2n+1} }{ (2n+1)! } = C _{0}\sum_{n=0}^{\infty}\frac{ (3x)^{n} }{ n! }+C _{1}\sum_{n=0}^{\infty}\frac{ (-3x)^{n} }{ n! }\]
\[= C _{0}e^{3x}+C_{1}e^{-3x}\]
where:
\[C_{0} + C_{1} = a_{0}\]and
\[C_{0}-C_{1} = \frac{ a_{1} }{ 3 }\]

Answer 2

I know how it becomes e^(3x) and all that jazz, thats the whole point anyway, but how it comes up with the substitution that eliminates the 2n, 2n+1 business and how the substitution turns out to be a plus and minus of C0 C1. Just not seeing where it came from.

Answer 3

Yes, backup indeed x_x

Answer 4

@Euler271 @Zarkon @tkhunny @douglaswinslowcooper @wio @mathmale Those I know they can help, hihihi because I don't know, either.

Answer 5

Lol, no worries xD

Answer 6

Oh well. This isnt for an assignment or something that needs to be rushed, Im just finishing off all my Diff Eq sections ive been going through. This is the last itty bit of it.

Answer 7

lol, talk about skipping steps!

sorry that i can't help you. good luck though

Answer 8

Yeah, thats straight to where the solution jumped to in the book :/ I mean, you can see the e^(3x) e^(-3x) junk hidden in there, but they sure got it out in a unique way. Wish there were more in between steps that I could provide, but thats all I was given -_-

Answer 9

Not that it would help, but the background of this problem: So, you start off with the homogenous DE y'' -9y = 0, which can be easily solved to be c1e^(3x) + c2e^(-3x).
you are supposed to show that you can get the same solution via power series. So this is basically the point along that power series path where I get stuck. I can get as far as I have typed out until it throws in the C0 C1 substitution. After that, I have no idea what was done. Perhaps there is a different way to get from the a0 a1/3 portion I have to the known solution?

Answer 10

Not sure if this will be of any help, but the first summation on the left is every n and the second one is every odd n and so if we combine them we can have (3x)^n / n! but we have to contend with (a1)/3 and (a0).

Answer 11

first summation i s **every even n**

Answer 12

Yeah. The strategy the text book gives is to separate them into their odd and even portions, supposedly so you can have two separate terms in which to apply the C constants to. So when I wrote out the series, I intentionally kept evens and odds apart. I suppose we could backtrack. Before I get the series, I have this:

\[a_{n+2} = \frac{ 9a_{n} }{ (n+1)(n+2) }\]

Answer 13

mathematical equivalent of yada, yada, yada... :)

Answer 14

Lol. Well, yada yada yada has to somehow get me c1e^(-3x) + c2e^(3x), lol.

Answer 15

c0 = 1/2(a0 + a1/3)
c1 = 1/2(a0 - a1/3)
Seems to suggest that each of the two summations on the left have to be split into two summations.

Answer 16

Thats what I was thinking. I tried using its answer tobacktrack and stopped, hoping there was something simpler. I could try going back to that idea and checking, though.

Answer 17

If you or anyone finds the answer please do post them here.
I am curious to find out and I will check back later.

Answer 18

Alrighty, will do.

Answer 19

\[e^{x} = \sum_{k = 0}^{\infty}\frac{ x^k }{ k! } = 1 + x + \frac{ x^2 }{ 2! } + \frac{ x^3 }{ 3! } + \frac{ x^4 }{ 4! } + ~...\]\[e^{3x} = ~~1 + 3x + \frac{ (3x)^2 }{ 2! } + \frac{ (3x)^3 }{ 3! } + \frac{ (3x)^4 }{ 4! } + ~...(1)\]\[e^{-3x} = 1 - 3x + \frac{ (3x)^2 }{ 2! } - \frac{ (3x)^3 }{ 3! } + \frac{ (3x)^4 }{ 4! } - ~...(2)\]Add (1) and (2) and divide by 2 you get every ODD term of the series e^(3x)
Subtract (2) from (1) and divide by 2 you get every EVEN term of the series e^(3x)\[a _{0} \sum_{n=0}^{\infty}\frac{ (3x)^{2n} }{ (2n)! } + \frac{ a _{1} }{ 3 }\sum_{n=0}^{\infty}\frac{ (3x)^{2n+1} }{ (2n+1)! } = \]a0 * (every odd term of the series e^(3x)) +
a1/3 * (every even term of the series e^(3x)) = \[a _{0}(\frac{ e^{3x} + e^{{-3x}} }{ 2 }) + \frac{a _{1}}{3}(\frac{ e^{3x} - e^{-3x} }{ 2 }) =\]\[e^{3x}(\frac{ a_{0} }{ 2 } + \frac{ a_{1} }{ 6 } ) + e^{-3x}(\frac{ a_{0} }{ 2 } - \frac{ a_{1} }{ 6 } ) = C_{0}e^{3x} + C_{1}e^{-3x}\]

Answer 20

PS: In series marked (2), the (-3x) is equivalently substituted by (+3x) when raised to even powers.

Odd terms have even powers and even terms have odd powers and that may be a bit confusing above.