Applying Inverse Laplace Transforms to f(s) = -5s/S^2+9

[Spoolx]

Homework Statement

f(s) = -5s/S^2+9

Homework Equations

I think
f(t) cosωt = f(s) s/s^2+ω^2

The Attempt at a Solution

ω=3

Answer
-5cos(3t)

Can anyone tell me if I did this correctly? I think I did but just want to make sure, if not can you tell me what I did wrong?

Thanks

 

[CAF123]

There is not really much of a problem statement there but, going by the title, I think you are after the inverse laplace transform of $f(s) = -5 \frac{s}{s^2+9}$. Yes, your result is correct. $$\mathcal{L}^{-1} f(s) \equiv Y(t) = -5 \cdot \mathcal{L}^{-1}\left( \frac{s}{s^2+9}\right) = -5 \cos 3 t$$

 

[Spoolx]

I just wanted to verify my answers, I don't have asolutions manual and want to make sure I am doing the problems correctly.

Thank you

 

[Ray Vickson]

I just wanted to verify my answers, I don't have asolutions manual and want to make sure I am doing the problems correctly.

Thank you

Your answer is wrong for what you WROTE, which was
f(s) = -\frac{5s}{s^2}+9
but it would be correct if you had written
f(s) = -\frac{5s}{s^2+9}
In text you would write this using parentheses: f(s) = -5s/(s^2+9). Such a simple step to avoid confusion!

 

[Spoolx]

I am sorry, the way you wrote it the second way is the way it was supposed to be written.. guess I need to learn to make proper equations.

Thanks again