Question 1

4. Compute the inverse Laplace transform of

$F(s)= \frac{1}{{s}^(4) } + \frac{3s}{{s}^(2) + 9 }$

$F(s)= (7)/(s) + \frac{6}{{s} + 4} - \frac{5}{ {s}^(2) + 16}$

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

$F(s)= \frac{s + 2}{ {s}^(2) (s + 1)(s - 1)}$

$F(s)= \frac{s + 1}{ {s}^(2) (s + 1)}$

$4. Compute the inverse Laplace transform of F(s)= \frac{1}{{s}^(4) } + \frac{3s}{{s-example-1$

Question 2

Answer:

See below for all answers and explanations

Explanation:

1st Problem

$F(s)=(1)/(s^4)+(3s)/(s^2+9)\\\\f(t)=(t^3)/(6)+3cos(3t)$

2nd Problem

$F(s)=(7)/(s)+(6)/(s+4)-(5)/(s^2+16)\\ \\f(t)=7+6e^(-4t)-(5)/(4)sin(4t)$

3rd Problem

$F(s)=(1)/((s-2)(s+1)(s-3))\\\\(1)/((s-2)(s+1)(s-3))=(A)/(s-2)+(B)/(s+1)+(C)/(s-3)\\ \\1=(s+1)(s-3)A+(s-2)(s-3)B+(s-2)(s+1)C$

$1=(-1+1)(-1-3)A+(-1-2)(-1-3)B+(-1-2)(-1+1)C\\1=12B\\(1)/(12)=B$

$1=(3+1)(3-3)+(3-2)(3-3)B+(3-2)(3+1)C\\1=4C\\(1)/(4)=C$

$1=(2+1)(2-3)A+(2-2)(2-3)B+(2-2)(2+1)C\\1=-3A\\-(1)/(3)=A$

$F(s)=(-(1)/(3))/(s-2)+((1)/(12))/(s+1)+((1)/(4))/(s-3)\\ \\f(t)=-(1)/(3)e^(2t)+(1)/(12)e^(-t)+(1)/(4)e^(3t)$

4th Problem

$F(s)=(s+2)/(s^2(s+1)(s-1))\\ \\(s+2)/(s^2(s+1)(s-1))=(A)/(s)+(B)/(s^2)+(C)/(s+1)+(D)/(s-1)\\\\s+2=s(s+1)(s-1)A+(s+1)(s-1)B+s^2(s-1)C+s^2(s+1)D\\ \\s+2=s^3A-sA+s^2B-B+s^3C-s^2C+s^3D+s^2D\\\\s+2=s^3(A+C+D)+s^2(B-C+D)-sA-B$