Question

Laplace Transformation

Answer 1

\[L[e^{2t}tcost]\]

Answer 2

so your initial function is :

(e^2t) x (t cos (t))...
yeah?

Answer 3

yeah

Answer 4

k have a look at this then
its from a table of laplace transforms

Answer 5

How to solve it without that?

Answer 6

page 3 and 4 give a pretty good explanation

Answer 7

for most exam type situations, you will be allowed to take in the table of standard transformations so as to attack the problem in a timely manner
if you wish to learn more tho I'd recommend:
http://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx
and
http://www.sosmath.com/diffeq/laplace/basic/basic.html

Answer 8

You can try to solve the integral,

\[F[s]=\int_0^\infty e^{(2-s)t}t\cos t dt\]
But, it should be better to solve first,
\[f[\alpha]=\int e^{\alpha t}\cos t dt\]
Because,
\[\int e^{\alpha t}t\cos t dt=\frac{\partial f}{\partial \alpha}\]
Following this philoshopy, first,
\[f[\alpha]=\int e^{\alpha t}\cos t dt=\frac{e^{\alpha t}}{1+\alpha^2}(\sin t+\alpha \cos t)\]
Then
\[\frac{\partial f}{\partial \alpha}=\frac{e^{\alpha t}((1+\alpha^2)(t(\sin t+\alpha\cos t)+\cos t)-2\alpha(\sin t+\alpha \cos t))}{(1+\alpha^2)^2}\]
Taking limits (exponentials vanish in infinite), you have,
\[F[\alpha]=\frac{\alpha^2-1}{(1+\alpha^2)^2}\]And
\[\alpha=2-s\]
You have,
\[F[s]=\frac{s^2-4s+3}{(s^2-4s+5)^2}\]

Answer 9

yeah I'd go with @John_ES 's method lynncake