I have to find a series solution to:
$$y'' - t^3 y =0$$
I go through the steps like so,
$$\sum_2 n(n-1)a_n x^{n-2} - \sum_0 a_n x^{n+3} = $$ $$\sum_0 (k+2)(k+1)a_{k+2} x^{k} - \sum_3 a_{k-3} x^{k} =$$
I am struggling to get the indices to math and I'm not sure of any methods to tackle this.
You are correct. Writing $$\sum_{n=0}^\infty n(n-1)a_n x^{n-2} - \sum_{n=0}^\infty a_n x^{n+3}=0$$ (I suggest you do not change the indices). Now, consider the above for power $n-2$ ; so $$n(n-1)a_{n}-a_{n-5}=0$$ (as you wrote). So, $$a_{n}=\frac{a_{n-5}}{n(n-1)}$$ with $a_2=a_3=a_4=0$. Since you face a second order differential equation, coefficients $a_0$ and $a_1$ are free (until you have conditions).