# Help with an easy Laplace Transform problem.

1. Sep 30, 2008

### wiz0r

Problem;

If, f(t) = d/dt [(e^-5t) (cos2t)]

Find F(s).

Eh, well, I don't really know what to do, can I get some pointers?

Am I supposed to integrate both sides, so I can get rid of d/dt, and then apply the integration property to find the answer?

Last edited: Sep 30, 2008
2. Sep 30, 2008

You should

• Know the transform of $$\cos 2t$$
• Have a theorem that tells you how to find the transform of a derivative of a function
• Know how multiplication by an exponential function influences the transform of a given function

Using these ideas, not necessarily in the order presented here, will get you through the problem.

3. Sep 30, 2008

### wiz0r

Anyway, first of all I made a typo on the initial equation. The correct f(t) is;

f(t) = d/dt [(e^-5t) (cos2t)]

Now, what I know is;

if f(t) = cos2t, then F(s) = s / (s^2 + 4)

also, if f(t) = (e^-5t) (cos2t) ,then F(s) = (s + 5) / [(s + 5)^2 +4]

Now, I know that;

df(t) / dt = sF(s) - f(0) and that f(0) = 1

Now, I will try to add everything together and find an answer..

4. Sep 30, 2008

### wiz0r

Umm, well;

I know that df(t) / dt = sF(s) - f(0) and that f(0) = 1

so,

L[f(t)] = s [(s + 5) / (s + 5)^2 +4] - 1

F(s) = [s(s+5) / (s+5)^2 + 4 ] - s

Am I correct?

Last edited: Sep 30, 2008
5. Oct 1, 2008

### Defennder

Where did this 's' come from? More importantly does the question say that f(0) = 1?

6. Oct 1, 2008

### wiz0r

You are correct, I can not say that f(0) = 1, also, that was another mistake(the s). But, well, I guess, that doesn't help me much. :x

Anyway, I found a ti89 program to find the transform. I know now that the answer is;

F(s) = -(5s + 29) / s^2 + 10s + 29

But, huh, no idea how to get there. It's 2:35 am, being doing all kind of problems with laplace transforms, and the first problem(this one), I cant solve it. Woot, for some reason I feel this problem or something really similar to this is coming on my exam tomorrow. >_>

7. Oct 1, 2008

### wiz0r

So, umm f(0) should be zero, right?

f(0) = d/dt ( cos0 ) = 0

Ok, now, I get;

F(s) = [s(s+5) / (s+5)^2 + 4 ]

F(s) = s(s+5) / s^2 + 10s + 29

Hey, well at least the denominator is correct. >.> Wonder what am I doing wrong.

8. Oct 1, 2008

### Defennder

You're right about that. f(0) = 1 by substitution. I thought we were solving a DE and the initial condition was omitted.

Your answer is correct. Just combine the -1 into the same fraction and you'll get what your program gave you.

9. Oct 1, 2008

### wiz0r

Alright. Thank you, Defender. Now I can sleep in peace.