Question

s2+15 X(s) (s2+5s+ 6) (s2 +9) Find: (a) Use Partial Fractions Decomposition to write the rational function as the sum of simpler expressions (b) Obtain the time-domain solution, x(t), by finding the inverse Laplace Transform of X(s) f(t)) had initial conditions, x(0) 0 and (c) Consider the inverse question, if the ODE (ä + ax + bx = 1, what was the input function in the time domain, f(t) (0)

Answer 1

S215 (s2 5s6) (s29)

2+15 (s2) (s3) (s29)

for partial fraction

s215 a3sa2 S2)(s3) (s2+9) s29 S 3 s2

(s215) (s2) (s 3) (s2 9) (s2)(s3) (s29) |a3s+a2) (s2) (s 3) (s2 9) 2 9 ao (s2) (s3) (s29) a1(s + 2) (s 3) (s29) + S2 S 3 S

s215 a(s3) (s29)a1 (s 2) (s s9)(asa2) (s+2) (s+3)

Plug in s 2 into the equation

19 13

Plug in s 3 into the equation

4 a1

19 s215 (s 3) (s9) 13 (s 3 2) (s9) (a3sa2) (s + 2) (s 3)

$s^{2}+15=s^{3}\left(a_{3}+\frac{5}{39}\right)+s^{2}\left(a_{2}+5 a_{3}+\frac{67}{39}\right)+s\left(5 a_{2}+6 a_{3}+\frac{15}{13}\right)+\left(6 a_{2}+\frac{513}{13}-24\right)$

$a_{3}=-\frac{5}{39}, a_{2}=-\frac{1}{13}$

19 5 S 39 13 13 $2+9 83

so the partial fraction is as given below,

$\frac{-5 s-3}{39\left(s^{2}+9\right)}+\frac{19}{13(s+2)}-\frac{4}{3(s+3)}$

now, the inverse of above partial form

$=L^{-1}\left\{-\frac{5 s}{39\left(s^{2}+9\right)}+\frac{19}{13(s+2)}-\frac{3}{39\left(s^{2}+9\right)}-\frac{4}{3(s+3)}\right\}$

$=-\frac{5}{39} L^{-1}\left\{\frac{s}{s^{2}+9}\right\}-\frac{3}{39} L^{-1}\left\{\frac{1}{s^{2}+9}\right\}+\frac{19}{13} L^{-1}\left\{\frac{1}{s+2}\right\}-\frac{4}{3} L^{-1}\left\{\frac{1}{s+3}\right\}$

$\begin{array}{l}{L^{-1}\left\{\frac{s}{s^{2}+9}\right\} : \cos (3 t)} \\ {L^{-1}\left\{\frac{1}{s^{2}+9}\right\} : \frac{1}{3} \sin (3 t)} \\ {L^{-1}\left\{\frac{1}{s+2}\right\} : e^{-2 t}} \\ {L^{-1}\left\{\frac{1}{s+3}\right\} : e^{-3 t}}\end{array}$

so final inverse Laplace is

$=-\frac{5}{39} \cos (3 t)-\frac{3}{39} \cdot \frac{1}{3} \sin (3 t)+\frac{19}{13} e^{-2 t}-\frac{4}{3} e^{-3 t}$