Question

Inverse Laplace Transforms

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

5. 1/(s+1)(s+2)(s^2+2s+10)

Answer 1

5.

$$\dfrac{1}{(s+1)(s+2)(s^2+2s+10)}=\dfrac{A}{s+1}+\dfrac{B}{s+2}$$

$$+\dfrac{Cs+D}{s^2+2s+10}=\dfrac{A(s+2)(s^2+2s+10)}{(s+1)(s+2)(s^2+2s+10)}$$

$$+\dfrac{B(s+1)(s^2+2s+10)}{(s+1)(s+2)(s^2+2s+10)}$$

$$+\dfrac{(Cs+D)(s+1)(s+2)}{(s+1)(s+2)(s^2+2s+10)}$$

$$s=-1:A(1-2+10)=1$$

$$s=-2:-B(4-4+10)=1$$

$$s=0: 20A+10B+2D=1$$

$$s=1:39A+26B+6C+6D=1$$

$$A=\dfrac{1}{9}, B=-\dfrac{1}{10}, D=-\dfrac{1}{9}, C=-\dfrac{1}{90}$$

$$L^{-1}(\dfrac{1}{(s+1)(s+2)(s^2+2s+10)})$$

$$=\dfrac{1}{9}L^{-1}(\dfrac{1}{s+1})-\dfrac{1}{10}L^{-1}(\dfrac{1}{s+2})$$

$$-\dfrac{1}{90}L^{-1}(\dfrac{s+1}{(s+1)^2+3^2})-\dfrac{1}{10}L^{-1}(\dfrac{1}{(s+1)^2+3^2})$$

$$=\dfrac{1}{9}e^{-t}-\dfrac{1}{10}e^{-2t}-\dfrac{1}{90}e^{-t}\cos(3t)-\dfrac{1}{30}e^{-t}\sin (3t)$$