Jtdylktuy

up vote 1 down vote favorite 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.. sequences-and-series