# Thread: laplace transforms are causing me trouble

1. ## laplace transforms are causing me trouble

I'm having a little trouble with these laplace transforms, don't know why... they look pretty simple from first glance was wondering if someone could give me some direction solving these.

1. $\displaystyle exp(-at+b)$

2. $\displaystyle sin(wt+b)$

3. $\displaystyle cosh^23t$

2. Originally Posted by action259
$\displaystyle exp(-at+b)$
I assume $\displaystyle a>0$.
---
$\displaystyle e^{-at+b}$
Now there is no rule for that. But you need to change it.
You can write.
$\displaystyle e^{-at}e^b$ because the product of these is adding exponents.

Remember that,
$\displaystyle \mathcal{L}e^{-at}=\frac{1}{s+a}$
And,
$\displaystyle \mathcal{L}\{\kappa f(x)\}=\kappa \mathcal{L}\{f(x)\}$

Since, $\displaystyle e^b$ is a KoNsTaNt.
We have,
$\displaystyle \frac{e^b}{s+a}$

3. Originally Posted by action259
$\displaystyle sin(wt+b)$
Did you Botox it?
$\displaystyle \sin \omega t\cos b+\cos \omega t\sin b$
And, $\displaystyle b$ is a KoNsTaNt.
Apply the rules,
$\displaystyle \mathcal{L}\{ \sin at\}=\frac{a}{s^2+a^2}$
$\displaystyle \mathcal{L}\{ \cos at\}=\frac{s}{s^2+a^2}$
Thus,
$\displaystyle \frac{\omega \cos b}{s^2+\omega^2}+\frac{s\sin b}{s^2+\omega^2}$

4. Originally Posted by action259
$\displaystyle cosh^23t$
The problem is that you have a square. However, you can eliminate it.
Use the following identity,
$\displaystyle \cosh^2 x=\frac{1+\cosh 2x}{2}$
This will eliminate the square and turn it into one of the forms on the table.
(If you want I can prove that identity. The reason why I am saying this is because my Calculus book does not contain this identity on its treatise on hyperbolic functions. Thus, I assumed might be unfamiliar).