SUBJECT: DIFFERENTIAL EQUATION **Problem #2:**Choose A or B, but not both.Given $ F(s) $, find $ f(t) = \mathcal{L}^{-1}\{F(s)\} $ using Laplace Transform Tables from Theorems 7.2 & 3. Simplify your answer.**Option A:**\[ F(s) = \frac{7s^2 + 10s + 19}{(s^2 + 7)(s+1)^2} \]**Option B:**\[ F(s) = \frac{5s + 1}{s^2 - 8s + 13} \]

Given: A) F(s)=7s2+10s+19s2+7s+12 Find f(t)=L-1Fs as follows. f(t)=L-1F(s)=L-17s2+10s+19s2+7s+12

Answered: Choose A or B, but not both. Given…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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SUBJECT: DIFFERENTIAL EQUATION

**Problem #2:**
Choose A or B, but not both.

Given $ F(s) $, find $ f(t) = \mathcal{L}^{-1}\{F(s)\} $ using Laplace Transform Tables from Theorems 7.2 & 3. Simplify your answer.

**Option A:**
\[ F(s) = \frac{7s^2 + 10s + 19}{(s^2 + 7)(s+1)^2} \]

**Option B:**
\[ F(s) = \frac{5s + 1}{s^2 - 8s + 13} \]

Expert Solution

Step 1

Given:

$F (s) = \frac{7 s^{2} + 10 s + 19}{(s^{2} + 7) {(s + 1)}^{2}}$

Find $f (t) = L^{- 1} \{F (s)\}$ as follows.

$\begin{array}{rcl} f (t) & = & L^{- 1} \{F (s)\} \\ = & L^{- 1} \{\frac{7 s^{2} + 10 s + 19}{(s^{2} + 7) {(s + 1)}^{2}}\} \end{array}$

Step 2

By partial fraction decomposition,

$\begin{array}{rcl} \frac{7 s^{2} + 10 s + 19}{(s^{2} + 7) {(s + 1)}^{2}} & = & \frac{A s + B}{s^{2} + 7} + \frac{C}{s + 1} + \frac{D}{{(s + 1)}^{2}} \\ = & \frac{(A s + B) {(s + 1)}^{2} + C (s + 1) (s^{2} + 7) + D (s^{2} + 7)}{(s^{2} + 7) {(s + 1)}^{2}} \end{array}$

Therefore,

$\begin{array}{rcl} 7 s^{2} + 10 s + 19 & = & (A s + B) {(s + 1)}^{2} + C (s + 1) (s^{2} + 7) + D (s^{2} + 7) \end{array}$ $\begin{array}{rcl} s \end{array} \begin{array}{rcl} \end{array}$ ³+7s+s²+7

For $s = - 1$ ,

$7 {(- 1)}^{2} + 10 (- 1) + 19 = D ({(- 1)}^{2} + 7) 7 - 10 + 19 = 8 D 8 D = 16 D = 2$

Step 3

Then we have,

$\begin{array}{rcl} 7 s^{2} + 10 s + 19 & = & (A s + B) {(s + 1)}^{2} + C (s + 1) (s^{2} + 7) + 2 (s^{2} + 7) \\ = & (A s + B) (s^{2} + 2 s + 1) + C (s^{3} + 7 s + s^{2} + 7) + 2 s^{2} + 14 \\ = & A s^{3} + 2 A s^{2} + A s + B s^{2} + 2 B s + B + C s^{3} + 7 C s + C s^{2} + 7 C + 2 s^{2} + 14 \\ = & s^{3} (A + C) + s^{2} (2 A + B + C + 2) + s (2 B + A + 7 C) + B + 7 C + 14 \end{array}$

Equating the coefficients of like terms, we get,

$\begin{array}{rcl} 2 A + B + C + 2 & = & 7 \\ 2 B + A + 7 C & = & 10 \\ B + 7 C + 14 & = & 19 \\ A + C & = & 0 \end{array}$

Solving the above system gives,

$\begin{array}{rcl} A & = & 0 \\ B & = & 5 \\ C & = & 0 \end{array}$

Step 4

Therefore,

$\begin{array}{rcl} \frac{7 s^{2} + 10 s + 19}{(s^{2} + 7) {(s + 1)}^{2}} & = & \frac{A s + B}{s^{2} + 7} + \frac{C}{s + 1} + \frac{D}{{(s + 1)}^{2}} \\ = & \frac{0 + 5}{s^{2} + 7} + \frac{0}{s + 1} + \frac{2}{{(s + 1)}^{2}} \\ = & \frac{5}{s^{2} + 7} + \frac{2}{{(s + 1)}^{2}} \end{array}$

Then,

$\begin{array}{rcl} f (t) & = & L^{- 1} \{F (s)\} \\ = & L^{- 1} \{\frac{7 s^{2} + 10 s + 19}{(s^{2} + 7) {(s + 1)}^{2}}\} \\ = & L^{- 1} \{\frac{5}{(s^{2} + 7)} + \frac{2}{{(s + 1)}^{2}}\} \\ = & L^{- 1} \{\frac{5}{(s^{2} + 7)}\} + L^{- 1} \{\frac{2}{{(s + 1)}^{2}}\} \\ = & 5 L^{- 1} \{\frac{1}{(s^{2} + {(\sqrt{7})}^{2})}\} + 2 L^{- 1} \{\frac{1}{{(s + 1)}^{2}}\} \\ = & \frac{5 \sin (\sqrt{7} t)}{\sqrt{7}} + 2 t e^{- t} \{Since L^{- 1} \{\frac{1}{(s^{2} + a^{2})}\} = \frac{s i n (a t)}{a} and L^{- 1} \{\frac{1}{{(s - a)}^{2}}\} = t e^{a t}\} \end{array}$

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