Question

Diff EQ: Question in the comments

Answer 1

\[F(S) = \int\limits_{0}^{\infty}e^{-st}f(t) dt\] show that \[F'(S) = L(-tf(t))\] where L is the Laplace transform

Answer 2

Have you tried taking the derivative of F(s) with respect to s?

Answer 3

\[\int\limits_{0}^{\infty}-te^{-st}\] like that?

Answer 4

what happen to f(t)?

Answer 5

i wasn't sure if that stayed in after taking derivative

Answer 6

the only thing that has the s variable is the e^(-st)

Answer 7

derivative of f(t) w.r.t s is 0?

Answer 8

yes but f(t) is a constant multiple so bring it down
and take derivative of e^(-st)

Answer 9

w.r.t. s of course

Answer 10

What do you mean by bring it down? I'm lost :/

Answer 11

do you know constant multiple rule?

Answer 12

\[\int\limits_{0}^{\infty}-te^{-st}f(t)dt\] like this?

Answer 13

\[\frac{d}{ds} (k \cdot g(s)) \\ =k \frac{d}{ds} g(s) \]

Answer 14

yes

Answer 15

ok

Answer 16

thats all i had to do?

Answer 17

well do you think you have proved what it asked you to prove?

Answer 18

what is \[L(-t f(t)) \text{ equal to ? }\]

Answer 19

Awesome! Thank you.

Answer 20

np