# McLaurin expansion

I've got an example and I wanted to know how it is expanded. Thanks for help.

$$h''(x)+2h'(x)+h(x)=0$$

$$h(x)=5xe^{-x}+2e^{-x}$$

Is converted to

$$h(x)={\sum _{n=0}^{\infty } \frac{5n{(-1)}^{n+1}x^n}{n!}}+{\sum _{n=0}^{\infty } \frac{2(-x)^n}{n!}}$$

I wanted to do this with

$$p(x)=7e^{x}+5e^{-x}$$

• The main point is knowing the expansion for $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. Then manipulate. The first term in the series for h(x) looks wrong. – herb steinberg Jan 19 '18 at 0:49

This is an answer: $$5 e^{-x} + 7xe^x = \sum_{n=0}^{\infty} \frac{5 (-x)^n + 7 x^{n+1}}{n!}$$