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17E

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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.
$17 . y^{''} + y^{'} - y = t^{3}; y (0) = 1, y^{'} (0) = 0$

The Initial value for $y'' + y' - y = t^{3}$ is $Y = \frac{s^{5} + s^{4} + 6}{s^{4} (s^{2} + s - 1)}$

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

Define $L \{y\} (s) = Y (s)$

Using the properties listed below, take the Laplace transform of the equation.

$L \{y'\} (s) = s L \{y\} (s) - y (0) L \{y''\} (s) = s^{2} L \{y\} (s) - s y (0) - y' (0) C \{t^{n}\} (s) = \frac{n!}{s^{n + 1}} L \{y''\} + L \{y'\} - L \{y\} = L \{t^{3}\}$

Substitute the properties into the equation.

$[s^{2} Y - s y (0) - y' (0)] + [s Y - y (0)] - Y = \frac{3!}{s^{4}}$

Substitute the initial conditions

$y (0) = 1 a n d y' (0) = 0 (s^{2} Y - s) + (s Y - 1) - Y = \frac{6}{s^{4}}$

Isolate the Y variable and solve:

$s^{2} Y + s Y - Y = \frac{6}{s^{4}} + s + 1 Y (s^{2} + s - 1) = \frac{s^{5} + s^{4} + 6}{s^{4}} Y = \frac{s^{5} + s^{4} + 6}{s^{4} (s^{2} + s - 1)}$

Therefore, the initial value for $y'' + y' - y = t^{3}$ is $Y = \frac{s^{5} + s^{4} + 6}{s^{4} (s^{2} + s - 1)}$

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.]

${te}^{2 t} \cos 5 t$

In Problems 29 - 32, use the method of Laplace transforms to find a general solution to the given differential equation by assuming $y (0) = a a n d y' (0) = b$ a and b are arbitrary constants.

$y'' - 5 y' + 6 y = - 6 t e^{2 t}$

In Problems 3–10, determine the Laplace transform of the given function. $e^{3 t} s i n 4 t$

In Problems 3-10, determine the Laplace transform of the given function.

${(t + 3)}^{2} - {(e^{t} + 3)}^{2}$

solve the given initial value problem using the method of Laplace transforms.

$y'' + 2 y' + 2 y = u (t - 2 π) - - u (t - 4 π); y (0) = 1, y' (0) = 1$

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