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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?

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    You want to find the $n$-th derivative of $f(x)$ at $0$. Once you know all of those, you will know the series. Start from $f''(x)+f(x)=e^{-x}$. Put $x=0$. So $f''(0)+f(0)=e^{-0}$, and therefore $f''(0)=1$. Next, differentiate both sides of given equation. We get $f'''(x)+f'(x)=-e^{-x}$. Put $x=0$. We get $f'''(0)=-3$. Differentiate again. We get $f^(4)(x)+f''(x)=e^{-x}$. Put $x=0$. We get $f^{(4)}(0)=?$. Differentiate again. A pattern will become apparent, maybe.2011-11-30
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    @Andre This was the most straightforward approach. Thanks!2011-12-01
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    Think about the procedure described by J.M. and Didier Piau. It is what you need to do in general. (The solve the DE then expand approach is ordinarily not feasible, so that answer is less relevant.)2011-12-02

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