Solving ODEs using power series Solve ODEs:
a) $t(2-t)x''-6(t-1)x'-4x=0, x(1)=1, x'(1)=0$
b) $x''-tx=0$
c) $x''-t^3x=0$
d) $(2+t^2)x''-tx'+4x=0$
So what I do is to let $x(t)= \sum_{i=0}^{\infty} a_it^i$ whence 
$x'(t)= \sum_{i=0}^{\infty} ia_it^{i-1}$ and $x''(t)= \sum_{i=0}^{\infty} i(i-1)a_it^{i-2}=\sum_{i=0}^{\infty}(i+2)(i+1)a_{i+2}t^i$
Then I substitute all of these to the left-hand side of the equation and try to make them "dependent of $t^i$".
For instance according to c) I've got $(i+2)(i+1)a_{i+2}=a_{i-3}$ and have no idea what to do.
According to b): $a_{i+2}=\frac{a_{i-1}}{(i+1)(i+2)}$
I've already seen examples on this forum, but when the formula was like $a_{i+2}=...a_i$ . We could then suppose that $a_0=0$ and $a_1=1$ or conversely and it was almost done, but don't know what to do in these cases.
Any help will be much appreciated. 
 A: Using b) as example, we get:
\begin{align}
\sum_{n=0}^{\infty}(n+2)\,(n+1)\,a_{n+2}\,t^n-t\sum_{n=0}^{\infty}a_{n}\,t^n& =0\Leftrightarrow\\
(0+2)(0+1)\,a_{0+2}\,t^0+\sum_{n=1}^{\infty}(n+2)\,(n+1)\,a_{n+2}\,t^n-\sum_{n=1}^{\infty}a_{n-1}\,t^n& =0\Leftrightarrow\\
 2\,a_2+\sum_{n=1}^{\infty}\left((n+2)\,(n+1)\,a_{n+2}\,-a_{n-1}\right)\,t^{n}&=0\Leftrightarrow\\
\end{align}
Now, we look at the terms in the summation. 
Because no term can nullify $2\,a_2$, it must be $0$:
\begin{align}
a_{2}=0
\end{align}
Since terms with different powers of $t$ also can't nullify each other for all $t$, their coefficients must all be $0$. Thus:
\begin{align}
(n+2)\,(n+1)\,a_{n+2}-a_{n-1}&=0\Leftrightarrow\\
a_{n+2}&=\frac{a_{n-1}}{(n+2)(n+1)}, \,\, n\geq1
\end{align}
This is a recurrence relation which gives you all of the $a_i$.
Together with the condition for $a_2$, you can get some of them: for instance, with $n=3$, you get $a_5=\frac{a_2}{20}=0$; with $n=6$ you get $a_8=\frac{a_5}{56}=0$ and so on. More generally, for $n=3p, \,p\in\mathbb N$, you get $a_{n+2}=a_{3p+2}=0$. This means there are no terms in $t^{3p}$ in the solution.
As for the rest of the $a_i$ (say, for $n=2,5,8,...$ and $n=1,4,7,...$), the recurrence relation still works. All you need is for one of them to be given - one for each of these two sets of $n$'s - to get all the others. 
How would you obtain them both? From the two initial conditions, which are missing here. Hence you end up with two "degrees of freedom", that is, (effectively) two undefined constants.
