How Do You Solve the Differential Equation tx'' + sint x = 0 Using Power Series?

[Kak-Hazhar]

Homework Statement

using the fact that \frac{sint}{t} = \sum^{}_{n=0}\frac{ (-1)^{n} t^{2n}}{(2n+1)!}

i wanted to write:

using that sint/t = ∑ (-1)ⁿ·t²ⁿ/(2n+1)!

find a power series solution of the equation tx'' +sint x = 0

Homework Equations

sorry i pushed the wrong button,

The Attempt at a Solution

I'm a beginner at LaTeX, a disaster at LaTeX

i get something like ∑n(n-1)a_nt^(n-2) +sint∑a_n = 0

 

[clamtrox]

Homework Statement

using the fact that \frac{sint}{t} = \sum^{}_{n=0}\frac{ (-1)^{n} t^{2n}}{(2n+1)!}

i wanted to write:

using that sint/t = ∑ (-1)ⁿ·t²ⁿ/(2n+1)!

find a power series solution of the equation tx'' +sint x = 0

Homework Equations

sorry i pushed the wrong button,

The Attempt at a Solution

I'm a beginner at LaTeX, a disaster at LaTeX

i get something like ∑n(n-1)a_nt^(n-2) +sint∑a_n = 0

You only need to use the tex-brackets at the beginning and at the end; don't clutter the formula with them. Also, you should use \sin (or\mathrm{sin}) instead of sin, makes the font right.

\frac{\sin t}{t} = \sum_{n=0}\frac{ (-1)^{n} t^{2n}}{(2n+1)!}

Okay, so how do find a power series solution? You write the DE as a sum over powers of t and then require that the coefficient is zero for all n. That gives you a recursion relation between the coefficients a_n.

i get something like ∑n(n-1)a_nt^(n-2) +sint∑a_n = 0

You're missing a t and a t^n there. Then write out the series expansion for sint/t and you can find a formal solution of the form

\sum_{n=0}^{\infty} A_n t^n = 0 where now A_n is something quite complicated, but it contains the recursion relation for the coefficients a_n. I am not sure if you can find an answer in terms of elementary functions; the form of the equation suggests to me that you can't.

 

[Kak-Hazhar]

You're missing a t and a t^n there. Then write out the series expansion for sint/t and you can find a formal solution of the form

\sum_{n=0}^{\infty} A_n t^n = 0 where now A_n is something quite complicated, but it contains the recursion relation for the coefficients a_n. I am not sure if you can find an answer in terms of elementary functions; the form of the equation suggests to me that you can't.

Well this is how far i get:
Using that

\frac{\sin t}{t} = \sum_{n=0}^{\infty}\frac{(-1)^{n} t^{2n} }{(2n+1)!}

for t≠0 i rewrite tx''+sint x =0 as

(\aleph); x'' + \frac{\sin t}{t}x = 0

Ansats: x(t) = \sum_{n=0}^{\infty}a_{n}t^{n}

then

x''(t) = \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2}

pluggin this and the ansats for x(t) into (\aleph)

i get

\sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n}t^{n}

here is where I'm not 100% sure.

i multiply the last two series getting

\sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n+1} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n}

which is equivalent to

\sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sin t\sum_{n=0}^{\infty}a_{n}

now what's next?

 

[clamtrox]

You cannot use the same indices with the two sums. Use for example m and n. What you should have then is something proportional to t^{2m+n}. Now you need to rescale the sums so that you will have something of the form \sum A_n t^n where then A_n will be zero for all n. So for example for the first sum you'd write
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} t^{n}.

 

[Kak-Hazhar]

..you should have then is something proportional to t^{2m+n}. Now you need to rescale the sums so that you will have something of the form \sum A_n t^n where then A_n will be zero for all n. So for example for the first sum you'd write
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} t^{n}.

<br /> \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{m=0}^{\infty}\frac{(-1)^{m}t^{2m} }{(2m+1)!}\sum_{n=0}^{\infty}a_{n}t^{n} <br />

\Leftrightarrow

\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}t^{n} + \sum_{^{m=0}_{n=0}}^{\infty}\frac{(-1)^{m}t^{2m+n} }{(2m+1)!}a_{n}

i know how to shift series and add them but now what?

sincerely

 

[clamtrox]

You shift the term on the right as well. Just notice when you do it: then you only sum over a finite number of m:s for each n.

 

[Kak-Hazhar]

You shift the term on the right as well. Just notice when you do it: then you only sum over a finite number of m:s for each n.

what's the substitution in the second series?

give me a hint

 

[clamtrox]

Since you want to have t^n there, you have to do n+2m -> n. Then a_n -> a_(n-2m), and the summation over m will be cut off by the fact that a_(-1) = 0. That's all there is to it.

 

[Kak-Hazhar]

Since you want to have t^n there, you have to do n+2m -> n. Then a_n -> a_(n-2m), and the summation over m will be cut off by the fact that a_(-1) = 0. That's all there is to it.

yes that is exactly what i tried but got confused.

i used 2m+n = k then m = 1/2·(k-n) and (2m+1)! --> (k+1-n)! and a_{n} \rightarrow a_{k-2m}

 

[clamtrox]

You don't want to replace any m:s, only n:s. So the only things you need to touch are a_n and t^n+2m. You should end up with

\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2} + \sum_{m=0}^{n/2} a_{n-2m} \frac{(-1)^m}{(2m+1)!}\right] t^n,

where the sum over m goes to m=n/2 for even values of n and to m=(n-1)/2 for odd values.

 

[Kak-Hazhar]

You don't want to replace any m:s, only n:s. So the only things you need to touch are a_n and t^n+2m. You should end up with

\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2} + \sum_{m=0}^{n/2} a_{n-2m} \frac{(-1)^m}{(2m+1)!}\right] t^n,

where the sum over m goes to m=n/2 for even values of n and to m=(n-1)/2 for odd values.

Danke Clamtrox