Answer to Question #233089 in Math for abi

Question

L^-1{ s+1 / (s²+2s+s)² }

Answer 1

$$\dfrac{s+1}{(s^2+2s+s)^2}=\dfrac{s+1}{(s^2+3s)^2}$$

$$=\dfrac{A}{s}+\dfrac{B}{s^2}+\dfrac{C}{s+3}+\dfrac{D}{(s+3)^2}$$

$$=\dfrac{As(s+3)^2+B(s+3)^2+Cs^2(s+3)+Ds^2}{(s^2+3s)^2}$$

$$=\dfrac{As^3+6As^2+9As+Bs^2+6Bs+9B+Cs^3+3Cs^2+Ds^2}{(s^2+3s)^2}$$

$s^3:A+C=0$

$s^2:6A+B+3C+D=0$

$s^1:9A+6B=1$

$s^0:9B=1$

$$A=\dfrac{1}{27}, B=\dfrac{1}{9}, C=-\dfrac{1}{27}, D=-\dfrac{2}{9}$$ $$L^{-1}(\dfrac{s+1}{(s^2+2s+s)^2})=\dfrac{1}{27}L^{-1}(\dfrac{1}{s})+\dfrac{1}{9}L^{-1}(\dfrac{1}{s^2})$$

$$-\dfrac{1}{27}L^{-1}(\dfrac{1}{s+3})-\dfrac{2}{9}L^{-1}(\dfrac{1}{(s+3)^2})$$

$$=\dfrac{1}{27}+\dfrac{t}{9}-\dfrac{1}{27}e^{-3t}-\dfrac{2}{9}te^{-3t}$$

$$L^{-1}(\dfrac{s+1}{(s^2+2s+s)^2})=\dfrac{1}{27}+\dfrac{t}{9}-\dfrac{1}{27}e^{-3t}-\dfrac{2}{9}te^{-3t}$$