Question

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)}`
​

Answer 1

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`

`\begin{cases} A + C + D = 0\\B - C + D = 0\\- A = 1\\- B = 2 \end{cases}`

`A+C+D=0\\(A+C+D=0)+(B-C+D)=0\\A+B+2D=0\\-1-2+2D=0\\-3+2D=0\\2D=3\\D=(3)/(2)`

`A+C+D=0\\-1+C+(3)/(2)=0\\ (1)/(2)+C=0\\ C=-(1)/(2)`

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

5th Problem

`F(s)=(s+1)/(s(s^2+1))\\\\(s+1)/(s(s^2+1))=(A)/(s)+(Bs+C)/(s^2+1)\\\\s+1=(s^2+1)A+s(Bs+C)\\\\s+1=s^2A+A+s^2B+sC\\\\s+1=s^2(A+B)+sC+A`

`\begin{cases} A + B = 0\\C = 1\\A = 1 \end{cases}`

`A+B=0\\1+B=0\\B=-1`

`F(s)=(1)/(s)+(-s+1)/(s^2+1)\\ \\F(s)=(1)/(s)+(1-s)/(s^2+1)\\\\F(s)=(1)/(s)+(1)/(s^2+1)+(-s)/(s^2+1)\\\\f(t)=1+sin(t)-cos(t)`