Question

Inverse Laplace Transforms

Find L^-1 {F(s)} when F(s) is given by

1. 1/(s+3)(s+7)

Answer 1

1.

$$\dfrac{1}{(s+3)(s+7)}=\dfrac{A}{s+3}+\dfrac{B}{s+7}$$

$$=\dfrac{As+7A+Bs+3B}{(s+3)(s+7)}$$

$$A+B=0$$

$$7A+3B=1$$

$$A=\dfrac{1}{4}, B=-\dfrac{1}{4}$$

$$L^{-1}(\dfrac{1}{(s+3)(s+7)})=\dfrac{1}{4}L^{-1}(\dfrac{1}{s+3})$$

$$-\dfrac{1}{4}L^{-1}(\dfrac{1}{s+7})=\dfrac{1}{4}e^{-3t}-\dfrac{1}{4}e^{-7t}$$