Question

4. Compute the inverse Laplace transform of
$F(s)= \frac{1}{{s}^{4} } + \frac{3s}{{s}^{2} + 9 }$
$F(s)= \frac{7}{s} + \frac{6}{{s} + 4} - \frac{5}{ {s}^{2} + 16}$
$F(s)= \frac{1}{(s - 2)(s + 1)(s - 3)}$
$F(s)= \frac{s + 2}{ {s}^{2} (s + 1)(s - 1)}$
$F(s)= \frac{s + 1}{ {s}^{2} (s + 1)}$
​

Answer 1

Answer:

See below for all answers and explanations

Step-by-step explanation:

1st Problem

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

2nd Problem

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

3rd Problem

$F(s)=\frac{1}{(s-2)(s+1)(s-3)}\\\\\frac{1}{(s-2)(s+1)(s-3)}=\frac{A}{s-2}+\frac{B}{s+1}+\frac{C}{s-3}\\ \\1=(s+1)(s-3)A+(s-2)(s-3)B+(s-2)(s+1)C$

$1=(-1+1)(-1-3)A+(-1-2)(-1-3)B+(-1-2)(-1+1)C\\1=12B\\\frac{1}{12}=B$

$1=(3+1)(3-3)+(3-2)(3-3)B+(3-2)(3+1)C\\1=4C\\\frac{1}{4}=C$

$1=(2+1)(2-3)A+(2-2)(2-3)B+(2-2)(2+1)C\\1=-3A\\-\frac{1}{3}=A$

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

4th Problem

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

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

$A+C+D=0\\(A+C+D=0)+(B-C+D)=0\\A+B+2D=0\\-1-2+2D=0\\-3+2D=0\\2D=3\\D=\frac{3}{2}$

$A+C+D=0\\-1+C+\frac{3}{2}=0\\ \frac{1}{2}+C=0\\ C=-\frac{1}{2}$

$F(s)=\frac{-1}{s}+\frac{-2}{s^2}+\frac{-\frac{1}{2}}{s+1}+\frac{\frac{3}{2}}{s-1}\\ \\f(t)=-1-2t-\frac{1}{2}e^{-t}+\frac{3}{2}e^{t}$

5th Problem

$F(s)=\frac{s+1}{s(s^2+1)}\\\\\frac{s+1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs+C}{s^2+1}\\\\s+1=(s^2+1)A+s(Bs+C)\\\\s+1=s^2A+A+s^2B+sC\\\\s+1=s^2(A+B)+sC+A$

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

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

$F(s)=\frac{1}{s}+\frac{-s+1}{s^2+1}\\ \\F(s)=\frac{1}{s}+\frac{1-s}{s^2+1}\\\\F(s)=\frac{1}{s}+\frac{1}{s^2+1}+\frac{-s}{s^2+1}\\\\f(t)=1+sin(t)-cos(t)$