Calculating the inverse Laplace transform of the following

Question

I'm given that $$F(s)=\frac{a+3s-6}{(s-2)^2+a^2}$$ and I want to calculate $$\mathcal{L}^{-1}(F(s))$$

My attempt at the solution is as follows:

First splitting $F(s)$ as follows $$F(s)=\frac{a}{(s-2)^{2}+a^{2}}+\frac{3(s-2)}{(s-2)^2+a^{2}}$$

Then I need to calculate $$\mathcal{L}^{-1}(\frac{a}{(s-2)^{2}+a^{2}})+\mathcal{L}^{-1}(\frac{3(s-2)}{(s-2)^{2}+a^{2}})$$

Now I know that $$\mathcal{L}(\sin(at))=\frac{a}{s^2+a^2}$$. However we have a $(s-2)^{2}$ on the denominator of $F(s)$, how would I deal with this to find the correct function?.Would someone please show me how this would be done?

Thanks for taking the time to read through this problem, any help would be appreciated.

Edit: I know the solution should be $$f(t)=e^{2t}sin(at)+3e^{2t}cos(at)$$ But i'm unsure where the exponential terms come from?

Answer 1

You can have a look here, the idea is

$$ L\left\{e^{at}\sin(bt) \right\} = \frac{b}{(s-a)^2+b^2} $$

and

$$ L\left\{e^{at}\cos(bt) \right\} = \frac{s-a}{(s-a)^2+b^2} $$

Answer 2

What does an $e^t$ factor do to a Laplace transform?

Suppose:

$\mathcal L\{f(t)\} = \int_0^{\infty} f(t)e^{-st}\ dt = F(s)$

And we want to find:$\mathcal L\{e^{at}f(t)\}$

$\int_0^{\infty} e^{at} f(t)e^{-st}\ dt\\ \int_0^{\infty} f(t)e^{-(s-a)t}\ dt\\ F(s-a)$

Coming the other way we have:

$\mathcal L\{\sin at\} = F(s) = \frac {a}{s^2 +a^2}\\ \mathcal L\{e^{2t}\sin at\} = F(s-2) = \frac {a}{(s-2)^2 +a^2}\\ \mathcal L\{\cos at\} = \frac {s}{s^2 +a^2}\\ \mathcal L\{e^{2t}\cos at\} = \frac {s-2}{(s-2)^2 +a^2}\\ $