Question

(Answer not yet found, work below) I'm finally getting to some convolution problems and am confused about how to approach them; first one posted below in a moment.

Answer 1

Use the Laplace Transform to solve the given Initial Value Problem:

\[y'+y=tsin(t), y(0)=0.\]

Answer 2

\[sY(s)-y(0)+Y(s)=(-1)^1\frac{d}{ds}\bigg(\mathcal{L}\left\{\sin(t)\right\}\bigg)= \frac{2s}{(s^2+1)^2}\]

Answer 3

\[Y(s)(s+1) = \frac{2s}{(s^{2}+1)^2}\]

Answer 4

\[Y(s)=\frac{2s}{(s^2+1)^1(s+1)}\]

Answer 5

@Kainui , this is where I run into trouble; I'm just not very familiar with how to do convolution, look at this, think of splitting up the functions, but otherwise have little idea of what to do.

Answer 6

back.

Answer 7

@agent0smith , I'm not sure if you're familiar with Differential Equations at all, but here's a problem dealing with convolution.

Answer 8

(brb if anybody stops by, short bike ride to use somewhere else's WiFi)

Answer 9

Use partial fraction to write
\[
\frac{2 s}{(s+1) \left(s^2+1\right)^2}=\frac{s-1}{2
\left(s^2+1\right)}+\frac{s+1}{\left(s^2+1\right)^2}-\frac{1}{2
(s+1)}
\]

Answer 10

I know I can do that, was curious if easier to do with convolution

Answer 11

Probably.

Answer 12

yes this will be easy if u use convolution:)
\[F(s)=\frac{ 2s }{ (s^2+1)(s+1) }\\so ~for ~convolution~we~take\\F_1(s)=\frac{ 2s }{ s^2+1 }\\and\\F_2(s)=\frac{ 1 }{ s+1 }\\\]
now \[f_1(t)=L^{-1}\frac{ 2s }{ s^2+1 }=2.\cos(t)\\and\\f_2(t)=L^{-1}\frac{ 1 }{ s+1 }=e^{-t}\\so\\ L^{-1}F(s)=f(t)=\int\limits_{0}^{t}f_1(u).f_2(t-u)du\\=\int\limits_{0}^{t}2\cos(u).e^{-(t-u)}du=2.e^{-t} \int\limits_{0}^{t}\cos(u).e^u.du\]
this integration is very easy now :)

Answer 13

@Mendicant_Bias

Answer 14

(Sorry about this,. didn't get a chance to look until now! Taking a look right now.)

Answer 15

\[2e^{-t}\int\limits_{0}^{t}\cos(u)e^{u}du=\sin(u)e^u-\int\limits_{0}^{t}\sin(u)e^udu;\]

Answer 16

\[-\int\limits_{0}^{t}\sin(u)e^{u}du=-\bigg(-\cos(u)e^{u}-\bigg[-\int\limits_{0}^{t}\cos(u)e^{u}du\bigg]\bigg)\]

Answer 17

\[2e^{-t}\int\limits_{0}^{t}\cos(u)e^{u}du-\int\limits_{0}^{t}\cos(u)e^{u}du=\sin(u)e^{u}+\cos(u)e^{u}\]

Answer 18

\[\int\limits_{0}^{t}\cos(u)e^{u}du=\frac{e^{t}[\sin(t)+\cos(t)]}{2e^{-t}}=\frac{{e^2t}[\sin(t)+\cos(t)]}{2}-\frac{1}{2}\]

Answer 19

Whoops, should be that whole thing not over two due to that outside factor.

Answer 20

I can't figure out where we went off, but we are pretty far off; the answer is \[y=-\frac{1}{2}e^{-t}+\frac{1}{2}\cos(t)-\frac{1}{2}t \cos(t)+\frac{1}{2}t \sin(t)\]

Answer 21

Ahh! @sidsiddhartha ! I see what it is, heh. It was probably my fault, I explicitly miswrote at the beginning (s^2+1)^(1) when it was squared, you took it, ran with that, and took the inverse laplace of it and came up with a cosine trig function, it was written as (s^2+1)^2 in all previous examples and was intended as such. Alright, let's see where we can go using convolution with that in mind.

Answer 22

\[F_{1}(s)=\frac{2s}{(s^2+1)^2}, \ \ \ F_{2}(s)=\frac{1}{s+1}\]

Answer 23

My PC is restarting in 5 minutes involuntarily, gonna go for a short time./

Answer 24

To check your answer
\[
\mathcal{L}_s^{-1}\left[\frac{2
s}{\left(s^2+1\right)^2
(s+1)}\right](t)=\\-\frac{1}{2} e^{-t}
\left(-e^t t \sin (t)-e^t \cos
(t)+e^t t \cos
(t)+1\right)=\\-\frac{e^{-t}}{2}+\frac{
1}{2} t \sin (t)-\frac{1}{2} t \cos
(t)+\frac{\cos (t)}{2}
\]

Answer 25

Alright, cool, PC is back, taking another shot at this.

Answer 26

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{2s}{(s^{2}+1)^{2}}\right\}=t \sin(t); \ \ \ \mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s+1}\right\}=e^{-t}\]

Answer 27

\[\int\limits_{0}^{t}\tau \sin(\tau)e^{-(t-\tau)}d \tau\]Is this the correct general setup for the convolution integral?

Answer 28

yes that looks fine

Answer 29

\[e^{-t}\int\limits_{0}^{t}\tau \sin(\tau)e^{\tau}d \tau\]

Answer 30

yeah but this integration will be tedious

Answer 31

I'm thinking integration by parts? It doesn't like it's going to work out in a totally manner either way, but I'm thinking group sin(tau) by itself, and then group tau e^tau as the second thing for u, dv, du, v.

Answer 32

@eliassaab , I'm guessing you used partial fractions? Thank you very much for the check.

Answer 33

yeah that should work :)

Answer 34

Alright, one sec.

Answer 35

(Nevermind, choosing to group things differently):\[u = \tau, \ du = d \tau; \ \ \ dv = \sin(\tau)e^{\tau}d \tau, \ v = \ ?\](Going to use an integral calculator to calculate sin(tau)e^(tau), at this point I'm running out of time, it's elementary, and I just need to focus on learning convolution)\[v = \frac{1}{2}e^{\tau}(\sin(\tau)-\cos(\tau))\]

Answer 36

\[e^{-t}\int\limits_{0}^{t}\tau \sin(\tau)e^{\tau}d \tau=\frac{1}{2}\tau e^{\tau}(\sin(\tau)-\cos(\tau))-\int\limits_{0}^{t}\sin{\tau} \ e^{\tau}d \tau\]

Answer 37

(Forgot the e^{-t} at beginning, will multiply in later)

Answer 38

\[\frac{1}{2} \tau e^{\tau}(\sin(\tau)-\cos(\tau))-\frac{1}{2}\int\limits_{0}^{t}e^{\tau}(\sin(\tau)-\cos(\tau))d \tau = \]
Alright, I follow how this is done so I'm going to just start anothetr convolution problem. Thank you, guys./