Question

Inverse Laplace Transforms

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

3. s-1/s^2 (s+3)

Answer 1

$$\dfrac{s-1}{s^2(s+3)}=\dfrac{A}{s+3}+\dfrac{B}{s}+\dfrac{C}{s^2}$$ $$=\dfrac{As^2+Bs^2+3Bs+Cs+3C}{(s+3)(s+7)}$$ $$A+B=0$$ $$3B+C=1$$ $$3C=-1$$

$$A=-\dfrac{4}{9}, B=\dfrac{4}{9}, C=-\dfrac{1}{3}$$

$$L^{-1}(\dfrac{1}{(s+3)(s+3)})=-\dfrac{4}{9}L^{-1}(\dfrac{1}{s+3})$$

$$+\dfrac{4}{9}L^{-1}(\dfrac{1}{s})-\dfrac{1}{3}L^{-1}(\dfrac{1}{s^2})$$

$$=-\dfrac{4}{9}e^{-3t}+\dfrac{4}{9}-\dfrac{1}{3}t$$