Ytukyg

We are given the IVP $x''(t) + 2tx'(t) − 8x(t) = 0$ subject to $x(0)=1$ and $x'(0)=0$.
I got:

$x(t)$ = $sum_{n=0}^∞=a_n.t^n$

$x'(t)= sum_{n=0}^∞=n.a_n.t^{n-1}$

$x''(t)= sum_{n=0}^∞=(n+1)(n+2).a_{n+2}.t^{n}$ using shift of sumbation.

Subbing in i got and simplifying i got:

$sum_{n=0}^∞=[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]t^n$.

And we want $[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]=0$

$a_{n+2} = (8-2n).a_n/(n+2)(n+1)$

We have $x(0)=1$ and $x'(0)=0$, so $a_o= 1$ and $a_1 = 0$, and we can get the rest of the solutions:

$a_2=4,a_3=0, a_4=4/3,a_5=0,a_6=0,a_7=0,a_8=0,a_9=0$

Have i attempted this correctly, and how do i get the final answers? Im not seeing the connection between the solutions..

The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:

$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$

although getting that from the series is not going to be easy.

EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.