Answer to Question #243672 in Calculus for JaytheCreator

Question

Find the inverse Laplace transform of the following (i) 𝑠 (𝑠+𝑎) 2+𝑏2, (ii) 𝑠 2−5 𝑠 3+4𝑠 2+3𝑠 ’ (iii) 3𝑠 (𝑠+1) 4, (iv) 1 (𝑠 2+1) 2. 

Answer 1

(i)

$$f(t)=L^{-1}\{F(s)\}$$

$$=L^{-1}\{\dfrac{s}{(s+a)^2+b^2}\}=L^{-1}\{\dfrac{s+a-a}{(s+a)^2+b^2}\}$$

$$=L^{-1}\{\dfrac{s+a}{(s+a)^2+b^2}\}-L^{-1}\{\dfrac{a}{(s+a)^2+b^2}\}$$

$$=e^{-at}\cos(bt)-\dfrac{a}{b}e^{-at}\sin(bt)$$

(ii)

$$f(t)=L^{-1}\{F(s)\}=L^{-1}\{\dfrac{s^2-5}{s^3+4s^2+3s}\}$$

$$s^3+4s^2+3s=s(s^2+4s+3)=s(s+1)(s+3)$$

$$\dfrac{s^2-5}{s^3+4s^2+3s}=\dfrac{A}{s}+\dfrac{B}{s+1}+\dfrac{C}{s+3}$$

$$=\dfrac{A(s+1)(s+3)+Bs(s+3)+Cs(s+1)}{s^3+4s^2+3s}$$

$$s=0: -5=3A=>A=-\dfrac{5}{3}$$

$$s=-1: -4=-2B=>B=2$$

$$s=-3: 4=6C=>C=\dfrac{2}{3}$$

$$f(t)=L^{-1}\{\dfrac{s^2-5}{s^3+4s^2+3s}\}$$

$$=-\dfrac{5}{3}L^{-1}\{\dfrac{1}{s}\}+2L^{-1}\{\dfrac{1}{s+1}\}+\dfrac{2}{3}L^{-1}\{\dfrac{1}{s+3}\}$$

$$=-\dfrac{5}{3}+2e^{-t}+\dfrac{2}{3}e^{-3t}$$

(iii)

$$f(t)=L^{-1}\{F(s)\}$$

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

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

$$=\dfrac{3}{2}t^2e^{-t}-\dfrac{1}{2}t^3e^{-at}$$

(iv)

$$f(t)=L^{-1}\{F(s)\}=L^{-1}\{\dfrac{1}{(s^2+1)^2}\}$$

$$=\dfrac{1}{2(1)^3}(\sin(1\cdot t)-1\cdot t\cos(1\cdot t))$$

$$=\dfrac{1}{2}\sin(t)-\dfrac{1}{2}t\cos(t)$$