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.
$\begin{array}{l} x''+y=1; & x(0)=1, & x^{'} (0)=1 \\ {x+y}^{''} =-1; & y(0)=1, & y^{'} (0)=-1 \end{array}$

The solution is

$x (t) = c o s t + e^{t} - 1, y (t) = c o s t - e^{t} + 1$

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:

$\begin{array}{l} x'' + y = 1; & x (0) = 1, & x^{'} (0) = 1 \\ x + y^{''} = - 1; & y (0) = 1, & y^{'} (0) = - 1 \end{array}$

Step 2: Apply the Laplace transform

Given initial value equations are,

$\begin{array}{l} x'' + y = 1; & x (0) = 1, & x^{'} (0) = 1.... (1) \\ x + y^{''} = - 1; & y (0) = 1, & y^{'} (0) = - 1.... (2) \end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{l} s^{2} x (s) - s x (0) - x^{'} (0) + y (s) = \frac{1}{s} \\ s^{2} x (s) - s - 1 + y (s) = \frac{1}{s} \\ s^{2} x (s) + y (s) = \frac{1 + s^{2} + s}{s} .... (3) \end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{l} x (s) + s^{2} y (s) - s y (0) - y^{'} (0) = \frac{1}{s} \\ x (s) + s^{2} y (s) - s + 1 = - \frac{1}{s} \\ x (s) + s^{2} y (s) = \frac{s^{2} - s - 1}{s} ..... (4) \end{array}$

Solving equation third and fourth we get

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

Using partial fraction we can write as

$y (s) = \frac{s}{s^{2} + 1} - \frac{1}{s - 1} + \frac{1}{s}$

Taking inverse Laplace transform we get

$y (t) = c o s t - e^{t} + 1$

Step 3: Solve the third and fourth equation

On solving equation (3) and (4) we, obtain

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

Using partial fraction we can write as

$x (s) = \frac{s}{s^{2} + 1} + \frac{1}{s - 1} - \frac{1}{s}$

Taking inverse Laplace transform we get

$x (t) = c o s t + e^{t} - 1$

Hence

$x (t) = c o s t + e^{t} - 1, y (t) = c o s t - e^{t} + 1$

Step 4: conclusion

The final solution is

$x (t) = c o s t + e^{t} - 1, y (t) = c o s t - e^{t} + 1$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

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.
$\begin{array}{l} x''+y=1; & x(0)=1, & x^{'} (0)=1 \\ {x+y}^{''} =-1; & y(0)=1, & y^{'} (0)=-1 \end{array}$

Step by Step Solution

Step 1: Given information

Step 2: Apply the Laplace transform

Step 3: Solve the third and fourth equation

Step 4: conclusion

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

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.x''+y=1;x(0)=1,x'(0)=1x+y''=-1;y(0)=1,y'(0)=-1

Step by Step Solution

Step 1: Given information

Step 2: Apply the Laplace transform

Step 3: Solve the third and fourth equation

Step 4: conclusion

Most popular questions for Math Textbooks

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Pure Maths

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.
$\begin{array}{l} x''+y=1; & x(0)=1, & x^{'} (0)=1 \\ {x+y}^{''} =-1; & y(0)=1, & y^{'} (0)=-1 \end{array}$