Hard power series problem

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.

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.