Question

te^(-t) sin2t can you solve this please using convolution transform in laplace

Answer 1

http://tutorial.math.lamar.edu/Classes/DE/ConvolutionIntegrals.aspx
there's a whole section devoted to laplace transform

Answer 2

@hartnn

Answer 3

i am working on it .

Answer 4

\(te^{-t}\sin 2t\) or \(te^{-t}\sin^2t\)?

Answer 5

the first one .. pls answer tnx

Answer 6

Do you want its Laplace transform or what?

Answer 7

convolution sir
or transform of integral

Answer 8

You haven't given anything to *solve*. What are you trying to determine?

Answer 9

te^-t sin2t

Answer 10

here we go !!!

Answer 11

sir sorry for over demanding ... did you use integration by parts?

Answer 12

no i just used defination of convolution theorem .

Answer 13

you need to find the inverse laplace transform of what @sami-21 got.

Answer 14

the convolution formula is this sir \[\int\limits_{0}^{t}\] f(t) g(t-T)dt

Answer 15

i just need the answer of the convolution sir b4 transforming it on laplace

Answer 16

yes you use it for inverse process .
if you have the inverse and you want to find the Laplace then go for that way .
but when you need in the forward way simply find the laplace of f(t) and g(t) Multiply thm .

Answer 17

ahhh sir this is last can u answer in inverse way tnx

Answer 18

convolution first sir then you inverse it

Answer 19

Oh. You want \(te^{-t}*\sin2t\)?

Answer 20

yes sir convolution first then you inverse laplace it tnx

Answer 21

\[\Large L (f*g)t=F(s)G(s)\]
here
\[\Large f(t)=te^{-t}\]
its Laplace is
\[\Large F(s)=\frac{1}{(s+1)^2}\]
\[\Large g(t)=\sin(2t)\]
\[\Large G(s)=\frac{2}{s^2+4}\]
just use the above mentioned convolution theorem to get the required result
just Multiply F(s)G(s) this is the answer .

Answer 22

Now I'm really confused. I don't understand the question.

Answer 23

$$\begin{align*}
\mathcal{L}\{te^{-t}*\sin2t\}&=\mathcal{L}\{te^{-t}\}\mathcal{L}\{\sin2t\}\\
&=\frac1{(s+1)^2}\cdot\frac2{s^2+4}\\
&=\frac2{(s+1)^2(s^2+4)}\\
\end{align*}$$

Answer 24

Use partial fraction decomposition to break into a sum (good luck solving the ugly system of linear equations) and take the Laplace transform of each of those parts.

Answer 25

sir i just knew that in using convolution u don't need to transform in f(s) right .. u will answer it by using only f(t), after u answer it by using this formula\[\int\limits_{0}^{t}f(t)g(t-T)dt \] and using integration by parts on it , then you can inverse transform i.. thats my point

Answer 26

That is the definition of convolution. You can evaluate that integral if you want.

Answer 27

pls integrate it sir by using by parts but first you follow the formula in convolution tnx

Answer 28

If you didn't want to use Laplace transforms why did you said you did?
$$\begin{align*}te^{-t}*\sin2t&=\int_0^t\tau e^{-(t-\tau)}\sin2\tau\,\mathrm{d}\tau\\&=\int_0^te^{-t}\tau e^\tau\sin2\tau\,\mathrm{d}\tau\\&=e^{-t}\int_0^t\tau e^\tau\sin2\tau\,\mathrm{d}\tau\end{align*}$$
Now, let \(u=\tau\) and \(\mathrm{d}v=e^\tau\sin2\tau\,\mathrm{d}\tau\);
thus \(\mathrm{d}u=\mathrm{d}\tau\) and \(v=\int e^\tau\sin2\tau\,\mathrm{d}\tau=e^\tau\sin2\tau+\frac12\int e^\tau\cos2\tau\,\mathrm{d}\tau=\dots=\frac15e^\tau(\sin2\tau-2\cos2\tau)\). I leave the rest to you!

Answer 29

tnhk you so much