-1
$\begingroup$

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

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

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – herb steinberg Jan 19 '18 at 0:49
-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I made a mistake. $\endgroup$ – Marianna Kalwat Jan 19 '18 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.