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18E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 350

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 $15 - 24$ , solve for $Y (s)$ , the Laplace transform of the solution $y (t)$ to the given initial value problem.
$y'' - 2 y' - y = e^{2 t} - e^{t}; y (0) = 1, y' (0) = 3$

The Initial value for $y'' - 2 y' - y = e^{2 t} - e^{t}$ is $Y (s) = \frac{s^{3} - 2 s^{2} - s + 3}{(s^{2} - 2 s - 1) (s - 2) (s - 1)}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the Laplace Transform

The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
$F (s) = \int_{0}^{\infty} f (t) e^{- st} t'$

Step 2: Determine the Laplace transform

Applying the Laplace transform and using its linearity we get

$L \{y'' - 2 y' - y\} = L \{e^{2 t} - e^{t}\} L \{y''\} - 2 L \{y'\} - L \{y\} = \frac{1}{s - 2} - \frac{1}{s - 1}$

Solve for the Laplace transform as:

$[s^{2} Y (s) - s y (0) - y' (0)] - 2 [s Y (s) - y (0)] - Y (s) = \frac{1}{(s - 2) (s - 1)} s^{2} Y (s) - s - 3 - 2 s Y (s) + 2 - Y (s) = \frac{1}{(s - 2) (s - 1)}$

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

Solve further as:

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

Therefore, the Initial value for $y'' - 2 y' - y = e^{2 t} - e^{t}$ is $Y (s) = \frac{s^{3} - 2 s^{2} - s + 3}{(s^{2} - 2 s - 1) (s - 2) (s - 1)}$

Most popular questions for Math Textbooks

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

$coshbt$

Question: In Problems $15 - 24$ , solve for $Y (s)$ , the Laplace transform of the solution $y (t)$ to the given initial value problem.

$y'' + 6 y = t^{2} - 1; y (0) = 0, y' (0) = - 1$

Determine the inverse Laplace transform of the given function.

$\frac{1}{s^{2} + 4 s + 8}$ .

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.

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

In Problems $15 - 24$ , solve for $Y (s)$ , the Laplace transform of the solution $y (t)$ to the given initial value problem.

$y'' - 2 y' + y = c o s t - s i n t; y (0) = 1, y' (0) = 3$

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