Calculating a Taylor Polynomial of a mystery function I need to calculate a taylor polynomial for a function $f:\mathbb{R} \to \mathbb{R}$ where we know the following
$$f\text{ }''(x)+f(x)=e^{-x} \text{ } \forall x$$ $$f(0)=0$$ $$f\text{ }'(0)=2$$
How would I even start?

Hint: Assume that
$$
f(x)=\sum\limits_{n\geqslant0}a_n\frac{x^n}{n!},\qquad g(x)=\sum\limits_{n\geqslant0}b_n\frac{x^n}{n!},
$$
and that $g=f''+f$.
Then
$$ 
f''(x)=\sum\limits_{n\geqslant0}a_{n+2}\frac{x^n}{n!},
$$
hence, for every $n\geqslant0$, $b_n$ is...
To do: Identify $b_n$ for every $n\geqslant0$ and translate the initial conditions on $f$ in terms of $a_0$ and $a_1$.
