Inverse Laplace Transform using partial fractions

Question

As a result of a Laplace transform, I found this result $$\frac{1}{s^2} \cdot \frac{2}{s^2 +4}$$ Now I used partial fractions: $2 = A(s^2+4) + B(s^2)$, giving: $(A+B)s^2 = 0$ and $4A=2$, Hence $A=\frac{1}{2}$, $B=-\frac{1}{2}$ So the fraction can be written as: $$\frac{1}{2}\frac{1}{s^2} - \frac{1}{2}\frac{1}{s^2+4}$$

But how to continue now? How do I get back to a function with "t"? Of course the first part is: $\frac{1}{2}t$, but I cannot see the second part.

Answer 1

Recall that

$$ \mathcal{L}\left\{\sin(bt)\right\}=\frac{b}{s^2+b^2} $$

therefore

$$ \mathcal{L}^{-1}\left\{\dfrac{1}{s^2+b^2}\right\}=\frac{1}{b}\sin(bt) $$

Answer 2

Now you take the inverse of each term. Remember that

$$ \mathcal{L}\{t \mu(t)\} = \frac{1}{s^2} $$

and

$$ \mathcal{L}\{\sin(\omega t)\mu(t)\} = \frac{\omega}{s^2+\omega^2} $$

where $\mu(t)$ is the unit step. Here's a list of functions and their transform

Answer 3

Notice that $$\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+4}\right\}=\sin(2t)$$

Now divide both sides by $2$ and you have your result.