Question

Can anyone explain Laplace transform for me? as well as an example? Thanks.

Answer 1

Like differentiation and indefinite integration, the Laplace transform changes one function into another function. It can be used to make some differential equations easier to solve.

Answer 2

i understand most of its concept, but i have problem solving differential equations

Answer 3

You take the Laplace transform of both sides, simplify a bit, and then take the inverse Laplace transform.

Answer 4

i just had a quiz, on which i had to transform \(f(t)=e^{at}\) using Laplace... can you help me with that? because I think i did it wrong

Answer 5

\[ \large
\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt \implies \mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st}e^{at}dt = \int_0^\infty e^{(a-s)t}dt
\]

Answer 6

This is just integration here.

Answer 7

Like you can do \(u=(a-s)t\quad du=(a-s)dt\). Since \(a\) and \(s\) are constants, it's simple.

Answer 8

\[\frac{1}{a-s}e^{(a-s)t}\] this is what i got... i don't know hoe to plug in the infinity though

Answer 9

After the \(u\) sub, you should get \[
\large
\frac{1}{a-s}\int_0^{-\infty}e^udu = \frac{1}{a-s}(e^{-\infty}-e^0) = \frac{1}{a-s}(0-1) = \frac{1}{s-a}
\]

Answer 10

how did you make the positive infinity into a negative one?

Answer 11

It is assumed that \(a<s\) so that \(\lim_{t \to\infty}u(t)= -\infty\)

Answer 12

ok this makes sense now. thank you so much!

Answer 13

can you do another example with me? we have to show what \(L(f'(t))\) is like and I got stuck after using the integration by parts

Answer 14

\[\large
\int_0^\infty e^{-st}f'(t)dt
\]\(u = e^{-st}\quad du=-se^{-st}\quad dv = f'(t)dt\quad v=f(t)\)

Answer 15

\[\large
=e^{-st}f(t)\bigg|_0^\infty - \int_0^\infty -se^{-st}f(t)dt = -f(0) + s \int_0^\infty e^{-st}f(t)dt \\= \mathcal{L}\{f(t)\}-f(0)
\]

Answer 16

thank you i understand it now. i think i messed up the \(du\) :(

Answer 17

The tricky thing here is that we assume that \(\lim_{t\to\infty}e^{-st}f(t)=0\)

Answer 18

that makes sense though. thanks again

Answer 19

Yeah it makes sense so long as \(f(t)\) is not an exponential function as well.