# Laplace transform

1. Feb 11, 2013

### kidi3

1. The problem statement, all variables and given/known data
Find the Laplace transform of the following function
t2 - 2t

3. The attempt at a solution
$$\begin{gathered} f\left( t \right) = {t^2} - 2t \hfill \\ l\left( {f\left( t \right)} \right) = F\left( s \right) = \int_0^\infty {{e^{ - st}} \cdot \left( {{t^2} - 2t} \right)ds} \hfill \\ \Leftrightarrow \hfill \\ = \int_0^\infty {{e^{ - st}}{t^2} - 2t{e^{ - st}}ds} \hfill \\ = \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\ = \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\ {\text{Integration by parts}} \hfill \\ \int_0^\infty {{e^{ - st}}{t^2}} ds \hfill \\ \int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\ v'\left( t \right) = {e^{ - st}} \Rightarrow v = \frac{{ - 1}}{s}{e^{ - st}} \hfill \\ u\left( t \right) = {t^2} \Rightarrow u'\left( t \right) = 2t \hfill \\ \int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\ \int_0^\infty {{e^{ - st}}{t^2}} ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \int_{}^{} {2t \cdot } \frac{{ - 1}}{s}{e^{ - st}}ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \frac{{ - 2}}{s}\int_{}^{} {t \cdot } {e^{ - st}}ds \hfill \\ \end{gathered}$$

am i doing it correctly?

Last edited: Feb 11, 2013
2. Feb 11, 2013

### Karnage1993

You're supposed to integrate with respect to $t$, not $s$! Switch the $ds$ with $dt$ and start over.

3. Feb 11, 2013

### kidi3

Ahh.. It makes sense now.

4. Feb 11, 2013

### kidi3

Could you tell me what i am doing wrong with this one..

Think i got it..

Last edited: Feb 11, 2013