Solve ODE using analytic solutions Let the following ODE:
$x'' + tx' + x = 0.$


*

*Find the general solution $x(t) = a_0 x_1(t) + a_1 x_2(t),$ with $a_0, a_1 \in  \mathbb{R}$ and $x_1(t), x_2(t)$ are $t$ power series convergent for $t\in\mathbb{R}$

*Show that $x_1(t)$ is the power series of the function $e^{-t^2/2}$. Use this to find another lineally independent solution of the given equation. Check that this second solution can be written as a power series like $x_2(t)$
My attemp:
The coefficients of my ODE are analytic, I suppose a solution with the following form:
$$x(t) = a_0 \sum_{k=0}^\infty b_k t^k + a_1 \sum_{k=0}^\infty c_k t^k$$
Then
$$x'(t) = a_0 \sum_{k=0}^\infty b_k k t^{k-1} + a_1 \sum_{k=1}^\infty c_k k t^{k-1}0$$
$$x''(t) = a_0 \sum_{k=2}^\infty b_k k(k-1) t^{k-2} + a_1 \sum_{k=2}^\infty c_k k (k-1) t^{k-2}$$
Putting this in the ODE we get
$$a_0 \sum_{k=2}^\infty b_k k(k-1) t^{k-2} + a_1  \sum_{k=2}^\infty c_k k (k-1) t^{k-2} + a_0 \sum_{k=1}^\infty b_k k t^k + a_1 \sum_{k=1}^\infty c_k k t^k  + a_0 \sum_{k=0}^\infty b_k t^k + a_1 \sum_{k=1}^\infty c_k t^k = 0$$
We get the following recurrence relation:
$$\begin{cases} k=0 \qquad a_0 b_2 2 + a_1 c_2 2 + a_0 b_0 + a_1 +c_0 \\ k \ge 1 \qquad a_0 b_{k+2} (k+2)(k+1) + a_1 c_{k+2} (k+2)(k+1) + a_0 k b_k + a_1 c_k k + a_0 b_k + a_1 c_k\end{cases}$$
Simplyfing:
$$ \begin{cases} a_0 (2 b_2 + b_0) + a_1 (2 c_2 +c_0) \\ a_0((k+2)b_{k+2} + b_k) + a_1((k+2)c_{k+2} + c_k) \end{cases}$$
I have tried to solve this recurrence relation with Mathematica using RSolve but it only gives me the solution $0$.
What have I done wrong?
 A: You did not do anything wrong. You just need not we the $a_0$ and $a_1$. These will show up later in the recursion.
Correction. The coefficients of my ODE are analytic, I suppose a solution with the following form:
$$x(t) = \sum_{k=0}^\infty b_k t^k$$
Then
$$tx'(t)=t\sum_{k=0}^\infty kb_k  t^{k-1}=\sum_{k=0}^\infty k\,b_k  t^k$$
$$x''(t)=\sum_{k=2}^\infty k(k-1)\,b_k\,t^{k-2}=\sum_{k=0}^\infty(k+1)(k+2)\,b_{k+2}\,t^k$$
Putting these in the ODE we get
$$
\sum_{k=0}^\infty(k+1)(k+2)b_{k+2}t^k+\sum_{k=0}^\infty kb_k t^k+\sum_{k=0}^\infty b_kt^k=0
$$
or
$$
\sum_{k=0}^\infty \big((k+1)(k+2)b_{k+2}+(k+1)b_k\big) t^k=0,
$$
or
$$
(k+1)(k+2)b_{k+2}+(k+1)b_k=0, \quad \text{for all $k\in\mathbb N$},
$$
which implies
$$
b_{k+2}=-\frac{1}{k+2}{b_k}.
$$
for $k\ge 2$. So, once you specify $b_0$, $b_1$ you have a unique solution $x=x(t;b_0,b_1)$. 
Thus the two solutions $x_1$, $x_2$ which span the solution space are
$$
x_1(t)=x(t;1,0), \,\,x_2(t)=x(t;0,1).
$$
So the first one is obtained by setting $b_0=1$ and $b_1=0$, while the second
one is obtained by setting $b_0=0$ and $b_1=1$.
In particular, the general solution $x=x(t;b_0,b_1)$ is expressed as
$$
x(t;b_0,b_1)=b_0x_1(t)+b_1x_2(t).
$$
More specifically,
$$
b_{2k}=-\frac{1}{2k}b_{2k-2}=\cdots=\frac{(-1)^k}{2^kk!}b_0,$$
$$
b_{2k+1}=-\frac{1}{2k+1}b_{2k-1}=\cdots=\frac{(-1)^k}{1\cdot3\cdots (2k+1)}b_1
=\frac{(-2)^k k!}{(2k+1)!}b_1
$$
Finally
$$
x_1(t)=\sum_{k=0}^\infty\frac{(-1)^k}{2^kk!}t^{2k}=\mathrm{e}^{-x^2/2},
$$
$$
x_2(t)=\sum_{k=0}^\infty\frac{(-2)^k k!}{(2k+1)!}t^{2k+1}=\mathrm{e}^{-t^2/2}\int_0^t\mathrm{e}^{s^2/2}\,ds.
$$
These two functions constitute a fundamental set of solutions of the equation, i.e., every other solution is a linear combination of them.
