Show the inverse laplace transform in general

Question

My problem is, show that:

$L^{-1}(\dfrac{s}{(s+a)(s+b)})=\dfrac{ae^{-at}-be^{-bt}}{a-b} $

MY ATTEMPT:

I think that i have to use the convulution theorem, so

$F(s)=\dfrac{1}{s+a}$ $G(s)=\dfrac{s}{s+b}$

then

$f(s)=L^{-1}F=e^{-at}$ $g(s)=L^{-1}G=-be^{-bt}+1$

but when i try to integrate, it doesn´t work, i dont get what the problem says. Can someone help me please? I´m kind of desperate. Thanks in advance

Answer 1

Let$$\frac{s}{(s+a)(s+b)} = \frac{x}{s+a}+\frac{y}{s+b}$$

At $s=-a$, $\frac{-a}{-a+b} = x$

At $s = -b$, $\frac{-b}{-b+a} = y$

So, $$\frac{s}{(s+a)(s+b)} = \frac{1}{a-b}\bigg[\frac{a}{s+a}-\frac{b}{s+b}\bigg]$$

So, $\mathcal{L}^{-1}\frac{s}{(s+a)(s+b)} = \frac{1}{a-b}\mathcal{L}^{-1}\bigg[\frac{a}{s+a}-\frac{b}{s+b}\bigg] = \frac{1}{a-b}(ae^{-at}-be^{-bt})$

Thus,

$$f(t) = \frac{ae^{-at}-be^{-bt}}{a-b}$$

Answer 2

$L^{-1}\{\dfrac{s}{(s+a)(s+b)}\}=L^{-1}\{\frac{1}{a-b}(\frac{a}{s+a}-\frac{b}{s+b})\}=\frac{1}{a-b} (ae^{-at}-be^{-bt})=\dfrac{ae^{-at}-be^{-bt}}{a-b}$

As $L(e^{at})=\frac{1}{s-a}$ and $L^{-1}(\frac{1}{s-a})=e^{at}$