Question

Use the Laplace Transform to solve the "integral equation"
y'(t)=sin(t)+the integral (from 0 to t) y(t-u)cos(u)du
with y(0)=0

Answer 1

Use the convolution theorem for the transform of the integral.
\[y'(t)=\sin t+\int_0^t (t-u)\cos u~du\]
where you would have
\[\mathcal{L}\left\{\int_0^t f(t-u)g(u)~du\right\}=\mathcal{L}\{f(t)\}\times\mathcal{L}\{g(t)\}\]

Answer 2

In this case, \(f(t-u)=t-u\) and \(g(u)=\cos u\), so you would have
\[\mathcal{L}\left\{\int_0^t (t-u)\cos u~du\right\}=\mathcal{L}\{t\}\times\mathcal{L}\{\cos t\}\]

Answer 3

there is a y under the integral too. will that change the process?

Answer 4

Sorry, must have missed it. The question is
\[y'(t)=\sin t+\int_0^t y(t-u)\cos u~du\]
right?

In this case, you can still use the convolution theorem, but set \(f(t-u)=y(t-u)\) (assuming that's written to mean \(y\) as a function of \(t-u\), and not the function \(y(t)\) times \(t-u\)), so you have
\[\begin{align*}
\mathcal{L}\left\{y'(t)\right\}&=\mathcal{L}\left\{\sin t\right\}+\mathcal{L}\left\{\int_0^t y(t-u)\cos u~du\right\}\\\\
s\mathcal{L}\left\{y(t)\right\}-y(0)&=\mathcal{L}\left\{\sin t\right\}+\mathcal{L}\left\{y(t)\right\}\mathcal{L}\left\{\cos t\right\}\\\\
\mathcal{L}\left\{y(t)\right\}\left(s-\mathcal{L}\left\{\cos t\right\}\right)&=\mathcal{L}\left\{\sin t\right\}
\end{align*}\]

Answer 5

ok i follow you. but this doesn't solve it does it? you would have the laplace of sin(t) divided by (s-laplace cos(t)).....that gives the laplace of y?

Answer 6

Right, you have to take the transforms of the sine and cosine, and you have an equaiton of the form \(\mathcal{L}\{y\}=\cdots\), then you'd solve for \(y\) by taking the inverse transform, \(\mathcal{L}^{-1}\left\{\mathcal{L}\{y\}\right\}=y=\mathcal{L}^{-1}\{\cdots\}\).

Answer 7

so I have L(y)= 1/s^3......so to take the inverse laplace transform of each side my final answer would be y=...something. wolfram showed me that the inverse laplace transform of 1/s^3 is t^2/2, but I have no idea how to actually find that. Can you show me how they got that answer?

Answer 8

wait, i am watching a khan academy video and I think since t^n=n!/s^n+1

Answer 9

It comes from the fact that
\[\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}\]

Answer 10

then 1/s^3 is the same as 2!/2s^2+1

Answer 11

which gives t^2/2 as the original function?

Answer 12

Yes

Answer 13

YESSSSSSS. so wait, is this all I have to do? I'm done?

Answer 14

Yes that's it :) The nice thing about solving ODES with initial conditions and the Laplace transform is no pesky \(C\)s to deal with.

Answer 15

So seriously I am done with this problem? That seemed so simple now that you showed the steps. What the heck lol! Thanks so much

Answer 16

Yes, you're done! You're welcome!

Answer 17

Would you be able to help me with something else really quick? It's another laplace transform. All i have to do is find the laplace transform of (1-e^-t)/t but this one isn't as nice as the last one in my opinion.

Answer 18

Wolfram says the laplace transform of that is ln(1+(1/s)) but I cannot figure out how they got it

Answer 19

Sure. For this one, you'll use the formula
\[\mathcal{L}\left\{\frac{1}{t}f(t)\right\}=\int_s^\infty F(u)~du\]
where \(F(u)=\mathcal{L}\{f(t)\}\). In this case, \(f(t)=1-e^{-t}\), so the transform would be
\[\mathcal{L}\left\{\frac{1}{t}(1-e^{-t})\right\}=\int_s^\infty \mathcal{L}_s\{1-e^{-t}\}~du\]
(The subscript on the Laplace operator is meant to denote that we're transforming from the \(t\) domain to \(s\).

Answer 20

Also, the formula was taken from this neat resource here: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

Answer 21

Slight correction, that subscript should be a \(u\), not an \(s\).

Answer 22

sweet! I think my problem is the 1-e^-t. if it was just the e part the it would just be 1/s+1 but thats not correct is it

Answer 23

\[\mathcal{L}\left\{1-e^{-t}\right\}=\mathcal{L}\{1\}-\mathcal{L}\{e^{-t}\}=\frac{1}{s}-\frac{1}{s+1}\]

Answer 24

which when integrated becomes..... ln(u)-ln(u+1)
from s to infinity?

Answer 25

Right.

Answer 26

Hmm that integral doesn't seem to converge...

Answer 27

so...do I take the limit of that then?

Answer 28

yeah something's not right

Answer 29

I have a class in 15 min, but I'll get back to this when I have a chance. (unless someone else can clear up the issue here)

Answer 30

thank you so much! I will wait for your help, you have helped so much in the past!!!

Answer 31

Alright, it doesn't look like we did anything wrong. I must have been thinking too far ahead of my actual computation.

You have
\[y=\int_s^\infty\left(\frac{1}{u}-\frac{1}{u+1}\right)~du\]
Like you mentioned, we'll need to consider the limit:
\[y=\lim_{t\to\infty}\int_s^t\left(\frac{1}{u}-\frac{1}{u+1}\right)~du\]
which gives
\[y=\lim_{t\to\infty}\bigg[\ln u-\ln (u+1)\bigg]_s^t\]
By the FTC we get
\[y=\lim_{t\to\infty}\bigg[\ln t-\ln (t+1)\bigg]-\ln s+\ln(s+1)=\ln\frac{s+1}{s}\]
as desired.

Answer 32

this looks great! thanks so much!!!! but what is the FTC?

Answer 33

Fundamental theorem of calc, the one that says if \(F(x)\) is the antiderivative of \(f(x)\), then
\[\int_a^b f(x)~dx=F(b)-F(a)\]