Question

find the inverse laplace transform of F (S)=10 (S+1)/S (S+2)(S+3)^3

Answer 1

$$\dfrac{10(s+1)}{s(s+2)(s+3)^3}=\dfrac{A}{s}+\dfrac{B}{s+2}$$

$$+\dfrac{C}{s+3}+\dfrac{D}{(s+3)^2}+\dfrac{E}{(s+3)^3}$$

$$10(s+1)=A(s+2)(s+3)^3+Bs(s+3)^3$$

$$+Cs(s+2)(s+3)^2+Ds(s+2)(s+3)$$

$$+Es(s+2)$$

$$s=0,10=A(2)(3)^3=>A=\dfrac{5}{27}$$

$$s=-2,-10=B(-2)(1)^3=>B=5$$

$$s=-3,-20=E(-3)(-1)=>E=-\dfrac{20}{3}$$

$$s^4, 0=A+B+C=>C=-\dfrac{140}{27}$$

$$s^3, 0=A(9+2)+9B+C(6+2)+D$$

$$=>D=-\dfrac{50}{9}$$

$$\dfrac{10(s+1)}{s(s+2)(s+3)^3}=\dfrac{5}{27}\cdot\dfrac{1}{s}+5\cdot\dfrac{1}{s+2}$$

$$-\dfrac{140}{27}\cdot\dfrac{1}{s+3}-\dfrac{50}{9}\cdot\dfrac{1}{(s+3)^2}-\dfrac{20}{3}\cdot\dfrac{1}{(s+3)^3}$$

$$L^{-1}(\dfrac{10(s+1)}{s(s+2)(s+3)^3})=\dfrac{5}{27}H(t)+5e^{-2t}$$

$$-\dfrac{140}{27}e^{-3t}-\dfrac{50}{9}te^{-3t}-\dfrac{10}{3}t^2e^{-3t}$$