Hard power series problem 
Consider the differential equation $$(1+t)y''+2y=0$$
with the variabel coefficient $(1+t)$, with $t\in \mathbb{R}$.
Set $y(t)=\sum_{n=0}^{\infty}a_nt^n$. What are the first 4 terms in the associated power series?


An Attempt
We write
$$(1+t)y''=\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}$$
$$2y=\sum_{n=0}^{\infty}2a_nt^{n} $$
Combining these three gives us
$$\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}+ \sum_{n=0}^{\infty}2a_nt^{n}$$
We now want all the powers to be $t^n$.
$$\sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=0}^{\infty}2a_nt^{n} $$
We finally want the index of summation to all be the same.
$$2a_0+2a_2+ \sum_{n=1}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=1}^{\infty}2a_nt^{n}$$
Which simplifies to
$$2a_0+2a_2+ \sum_{n=1}^{\infty}\big[a_{n+2}(n+2)(n+1)+a_{n+1}n(n+1)+2a_n \big]t^n$$
Clearly, I'm not getting anywhere close to any of the answers given. I just can't find what I'm doing wrong. I have also tried with Maple, but that didn't give me any real result either.
I hope someone can help me out.
 A: Take $t=0$. You get from the ODE $y^{\prime \prime}(0)+y(0)=0$.
And answer b) is the only valid option providing that at least one is valid.
A: If
$$y(t)=\sum_{n=0}^{\infty}c_n t^n$$
then
$$y''(t)=\sum_{n=2}^{\infty}n(n-1)c_n t^{n-2}$$
which is equivalent to
$$y''(t)=\sum_{n=0}^{\infty}(n+1)(n+2)c_{n+2}t^n$$
so
\begin{align*}
(1+t)y''(t) &= y''(t)+ty''(t)\\
&=\sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}t^n\right]+t\sum_{n=0}^{\infty}(n+1)(n+2)c_{n+2}t^n\\
&= \left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)+t\left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)\\
&= \left(2c_2+2\cdot 3c_3 t+3\cdot 4c_4 t^2+\cdots\right)+\left(2c_2 t+2\cdot 3c_3 t^2+3\cdot 4c_4 t^3+\cdots\right)\\
&= \left(1\cdot 2c_2+0\cdot 1c_1\right)+\left(2\cdot 3c_3+1\cdot 2c_2\right)t+\left(3\cdot 4c_4+2\cdot 3c_3\right)t^2+\cdots\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}\right]t^n
\end{align*}
It follows that
\begin{align*}
0 &= (1+t)y''(t)+2y(t)\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}\right]t^n+\sum_{n=0}^{\infty}2c_n t^n\\
&= \sum_{n=0}^{\infty}\left[(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}+2c_n\right]t^n
\end{align*}
and upon equating coefficients,
$$(n+1)(n+2)c_{n+2}+n(n+1)c_{n+1}+2c_n=0$$
No initial conditions were given, so $c_0$ and $c_1$ can be chosen arbitrarily. The recurrence relation above then gives
\begin{align*}
c_2 &= -c_0\\
c_3 &= \frac{c_0-c_1}{3}\\
&\text{ }\vdots
\end{align*}
Thus,
$$y(t)=c_0+c_1 t-c_0t^2+\frac{c_0-c_1}{3}t^3+\cdots$$
so (d) is the correct answer.
Edit: I’m apparently blind! While writing this answer, I didn’t notice that (d) is the exact same series I derived. In light of this observation (thank you Carl!), I’ve edited out the last part of my original response, where I claimed that (b) was the correct answer.
