Help me fast so that I will give good rating. - 7s4s +7Q5.A.i. Find the inverse Laplace transform of F(s)%=(s -4) - 21 (s+1(2s+3)(3s+5)ii. Find the Laplace transform of f(t)= te " cosh5t.41d'yB.i. Solve the boundary value problemdt?+2y 0, with y(0) =1 & y(7)= 0.dyii. By using Laplace transform, solve the differential equation+10y = e", y(0)= 25.dt

Answered: - 7s 4s +7 Q5.A.i. Find the inverse…

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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Help me fast so that I will give good rating.

- 7s
4s +7
Q5.A.i. Find the inverse Laplace transform of F(s)%=
(s -4) - 21 (s+1(2s+3)(3s+5)
ii. Find the Laplace transform of f(t)= te " cosh5t.
41
d'y
B.i. Solve the boundary value problem
dt?
+2y 0, with y(0) =1 & y(7)= 0.
dy
ii. By using Laplace transform, solve the differential equation
+10y = e", y(0)= 25.
dt

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Part 1-

Given:

Given $F (s) = \frac{- 7 s}{{(s - 4)}^{2} - 21} - \frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}$

We want to find inverse Laplace transform of $F (s) = \frac{- 7 s}{{(s - 4)}^{2} - 21} - \frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}$

Calculation:

$\begin{array}{rcl} L^{- 1} \{F (s)\} & = & L^{- 1} \{\frac{- 7 s}{{(s - 4)}^{2} - 21} - \frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}\} \\ L^{- 1} \{F (s)\} & = & L^{- 1} \{\frac{- 7 s}{{(s - 4)}^{2} - 21}\} - L^{- 1} \{\frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}\} \\ . . . . . . . . . . . . . . . . . . . . . . . . . . (1) \end{array}$

Now consider

$\begin{array}{rcl} L^{- 1} \{\frac{- 7 s}{{(s - 4)}^{2} - 21}\} & = & L^{- 1} \{\frac{- 7 s + 28 - 28}{{(s - 4)}^{2} - 21}\} \\ = & L^{- 1} \{\frac{- 7 s + 28}{{(s - 4)}^{2} - 21}\} - L^{- 1} \{\frac{28}{{(s - 4)}^{2} - 21}\} \\ = & - 7 L^{- 1} \{\frac{s - 4}{{(s - 4)}^{2} - 21}\} - 28 L^{- 1} \{\frac{1}{{(s - 4)}^{2} - 21}\} \\ = & - 7 e^{4 t} L^{- 1} \{\frac{s}{s^{2} - 21}\} - e^{4 t} L^{- 1} \{\frac{28}{s^{2} - 21}\} \\ L^{- 1} \{\frac{- 7 s}{{(s - 4)}^{2} - 21}\} & = & - 7 e^{4 t} \cos h (\sqrt{21} t) - \frac{4 \sqrt{7}}{3} e^{4 t} [\sin h (\sqrt{21} t)] \\ . . . . . . . . . . . . . . . . . . . . . . . . . (2) \end{array}$

Now,

$\begin{array}{rcl} L^{- 1} \{\frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}\} & = & L^{- 1} \{\frac{3}{2 (s + 1)} - \frac{4}{(2 s + 3)} + \frac{3}{2 (3 s + 5)}\} \\ = & \frac{3}{2} L^{- 1} \{\frac{1}{(s + 1)}\} - L^{- 1} \{\frac{4}{(2 s + 3)}\} + \frac{1}{2} L^{- 1} \{\frac{3}{(3 s + 5)}\} \\ = & \frac{3}{2} L^{- 1} \{\frac{1}{(s + 1)}\} - 2 L^{- 1} \{\frac{2}{(2 s + 3)}\} + \frac{1}{2} L^{- 1} \{\frac{3}{(3 s + 5)}\} \\ = & \frac{3}{2} L^{- 1} \{\frac{1}{(s + 1)}\} - 2 L^{- 1} \{\frac{1}{(s + 3 / 2)}\} + \frac{1}{2} L^{- 1} \{\frac{1}{(s + 5 / 3)}\} \\ L^{- 1} \{\frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}\} & = & \frac{3}{2} e^{- t} - 2 e^{- 3 t / 2} + \frac{1}{2} e^{- 5 t / 3} \\ . . . . . . . . . . . . . . . . . . . . . . . . (3) \end{array}$

Therefore,

$\begin{array}{rcl} L^{- 1} \{F (s)\} & = & L^{- 1} \{\frac{- 7 s}{{(s - 4)}^{2} - 21}\} - L^{- 1} \{\frac{4 s + 7}{(s + 1) (2 s + 3) (3 s + 5)}\} \\ = & [- 7 e^{4 t} \cos h (\sqrt{21} t) - \frac{4 \sqrt{7}}{3} e^{4 t} [\sin h (\sqrt{21} t)]] - [\frac{3}{2} e^{- t} - 2 e^{- \frac{3 t}{2}} + \frac{1}{2} e^{- \frac{5 t}{3}}] \\ L^{- 1} \{F (s)\} & = & - 7 e^{4 t} \cos h (\sqrt{21} t) - \frac{4 \sqrt{7}}{3} e^{4 t} [\sin h (\sqrt{21} t)] - \frac{3}{2} e^{- t} + 2 e^{- \frac{3 t}{2}} - \frac{1}{2} e^{- \frac{5 t}{3}} \end{array}$

Answer:

$\begin{array}{rcl} L \end{array} \begin{array}{rcl} \end{array}$ ^-1Fs=-7e^4tcosh21t-473e^4tsinh21t-32e^-t+2e^-3t2-12e^-5t3

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