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Q8E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 413

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.
$\begin{array}{l} D [x] + y = 0; x (0) = 7 / 4, \\ 4 x + D [y] = 3; y (0) = 4 \end{array}$

The solution is $\begin{array}{l} x (t) = \frac{3}{4} + \frac{3 e^{- 2 t}}{2} - \frac{1 e^{2 t}}{2}, y (t) = 3 e^{- 2 t} + e^{2 t} \end{array}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

The differential equations are given as:

$\begin{array}{l} D [x] + y = 0; x (0) = 7 / 4, \\ 4 x + D [y] = 3; y (0) = 4 \end{array}$

Step 2: Apply the Laplace transform

Given initial value equations are,

$\begin{array}{l} D [x] + y = 0; x (0) = 7 / 4, \\ ∴ x^{'} + y = 0.... (1) \\ 4 x + D [y] = 3; y (0) = 4 \\ ∴ 4 x + y^{'} = 3..... (2) \end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{l} s x (s) - x (0) + y (s) = 0 \\ s x (s) - \frac{7}{4} + y (s) = 0 \\ y (s) = \frac{7}{4} - s x (s) ..... (3) \end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{r} 4 x (s) + s y (s) - y (0) = \frac{3}{s} \\ 4 x (s) + s y (s) - 4 = \frac{3}{s} ... (4) \end{array}$

Putting equation third into fourth we get

$\begin{array}{l} 4 x (s) + s [\frac{7}{4} - s x (s)] - 4 = \frac{3}{s} \\ [4 - s^{2}] x (s) = \frac{12 + 16 s - 7 s^{2}}{4 s} \\ x (s) = \frac{7 s^{2} - 16 s - 12}{4 s (s^{2} - 4)} \end{array}$

Using partial fraction we can write as

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

Taking inverse Laplace transform we get

$x (t) = \frac{3}{4} + \frac{3 e^{- 2 t}}{2} - \frac{1 e^{2 t}}{2}$

Since equation first is,

$\begin{array}{l} x^{'} + y = 0 \\ - 3 e^{- 2 t} - e^{2 t} + y (t) = 0 \\ y (t) = 3 e^{- 2 t} + e^{2 t} \end{array}$

Hence

$\begin{array}{l} x (t) = \frac{3}{4} + \frac{3 e^{- 2 t}}{2} - \frac{1 e^{2 t}}{2}, y (t) = 3 e^{- 2 t} + e^{2 t} \end{array}$

Step 3: Conclusion

The final solution is

$\begin{array}{l} x (t) = \frac{3}{4} + \frac{3 e^{- 2 t}}{2} - \frac{1 e^{2 t}}{2}, y (t) = 3 e^{- 2 t} + e^{2 t} \end{array}$

Most popular questions for Math Textbooks

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