Answer to Question #280082 in Calculus for jay

Question

a)     Determine the Laplace transforms of the following functions

(i) f(t) = t³-2t

(ii) f(t) = sin3t-e^2t

(iii) f(t) = e^2t sinh4t

b) Find the Laplace transform of the following

(i) f(t) = 3e^-4t- 5e^4t

(ii) f(t) = t sin 3t + cos 4t

find the inverse transform of

(i) F (s) = 2/3-1/2s-3

(ii) F (s) = 5s-8/s(s-4), using partial fractions.

Answer 1

a)     

(i) $f(t) = t^3-2t$

$$F(s)=L\{t^3-2t\}$$

$$=\dfrac{6}{s^4}-\dfrac{2}{s^2}$$

ii) $f(t) = \sin(3t)-e^{2t}$

$$F(s)=L\{\sin(3t)-e^{2t}\}$$

$$=\dfrac{3}{s^2+9}-\dfrac{1}{s-2}$$

iii) $f(t) = e^{2t}\sinh(4t)$

$$F(s)=L\{e^{2t}\sinh(4t)\}$$

$$=\dfrac{4}{(s-2)^2-16}$$

b)

(i) $f(t) = 3e^{-4t}- 5e^{4t}$

$$F(s)=L\{3e^{-4t}- 5e^{4t}\}$$

$$=\dfrac{3}{s+4}-\dfrac{5}{s-4}$$

(ii) $f(t) = t\sin(3t)+\cos(4t)$

$$F(s)=L\{ t\sin(3t)+\cos(4t)\}$$

$$=\dfrac{6s}{(s^2+9)^2}-\dfrac{s}{s^2+16}$$

c)

(i) $F (s) =\dfrac{2}{3}-\dfrac{1}{2s-3}$

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

$$=\dfrac{2}{3}\delta(t)-\dfrac{1}{2}e^{3t/2}$$

(ii) $F (s) =\dfrac{5s-8}{s(s-4)}$

$$\dfrac{5s-8}{s(s-4)}=\dfrac{A}{s}+\dfrac{B}{s-4}=\dfrac{A(s-4)+Bs}{s(s-4)}$$

$$s=0:-4A=-8=>A=2$$

$$s=0:4B=12=>B=3$$

$$f(t)=L^{-1}\{\dfrac{5s-8}{s(s-4)}\}=L^{-1}\{\dfrac{2}{s}\}+L^{-1}\{\dfrac{3}{s-4}\}$$

$$=2+3e^{4t}$$