Inverse laplace transform of $\frac{1}{(s+a)(s+b)}$

Question

Been trying to find the inverse laplace transform of $$\frac{1}{(s+a)(s+b)}$$ where $a$ and $b$ are constants.

Most obvious thing to do is try a partial fraction decomposition, but it just becomes a mess with the constants.

Answer 1

The partial fraction decomposition isn't so difficult (provided $a\not=b$). Begin by writing $$ \frac{1}{(s+a)(s+b)}=\frac{c_1}{s+a}+\frac{c_2}{s+b}. $$ Then, clear your fractions by multiplying through by $(s+a)(s+b)$ to get $$ 1=c_1(s+b)+c_2(s+a). $$ Then, you get $$ 1=(c_1+c_2)s+(c_1b+c_2a). $$ Equating coefficients of $s$, we get that \begin{align} c_1+c_2&=0\\ c_1b+c_2a&=1. \end{align} The first equation tells you that $c_2=-c_1$, while the second gives us $$ c_1b-c_1a=1 $$ or that $$ c_1=\frac{1}{b-a}. $$ Therefore, $$ \frac{1}{(s+a)(s+b)}=\frac{1}{b-a}\cdot\frac{1}{s+a}-\frac{1}{b-a}\cdot\frac{1}{s+b}. $$ Remember that when you take the inverse Laplace of these, $\frac{1}{b-a}$ is a constant, so it just factors through.

If $a=b$, then there are direct formulas for the inverse Laplace formula.

Answer 2

Here’s a way to make partial fractions quicker,

$$\frac{1}{(s+a)(s+b)}=\frac{c_1}{s+a}+\frac{c_2}{s+b}$$

For $s \neq -a$ and $s \neq -b$.

Multiplying both sides by $s+b$,

$$\frac{1}{(s+a)}=c_1\frac{s+b}{s+a}+c_2$$

Now take the limit as $s \to -b$.

$$\frac{1}{a-b}=c_2$$

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Try the same thing but this time multiplying by $s+a$.