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3E

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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 $1 - 14$ , solve the given initial value problem using the method of Laplace transforms.
$3 . y'' + 6 y' + 9 y = 0; y (0) = - 1, y' (0) = 6$

The Initial value for $y'' + 6 y' + 9 y = 0$ is $y (t) = - e^{- 3 t} + 3 t e^{- 3 t}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define 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 initial value of Laplace transform

Applying the Laplace transform and using its linearity we get

$L \{y'' + y' + 9 y\} = 0 L \{y''\} + 6 L \{y'\} + 9 L \{y\} = 0 [s^{2} Y (s) - s y (0) - y' (0)] + 6 [s Y (s) - y (0)] + 9 Y (s) = 0$

Solve for the Laplace transform as:

$s^{2} Y (s) + s - 6 + 6 [s Y (s) + 1] + 9 Y (s) = 0 s^{2} Y (s) + 6 s Y (s) + 9 Y (s) = - s (s^{2} + 6 s + 9) Y (s) = - s Y (s) = \frac{- s}{s^{2} + 6 s + 9}$

Using partial fractions solve as:

$\frac{- s}{s^{2} + 6 s + 9} = \frac{- s}{{(s + 3)}^{2}} = \frac{A}{s + 3} + \frac{B}{{(s + 3)}^{2}} - s = A (s + 3) + B$

Using $s = - 3, 0$ , , respectively, gives

$s = - 3 : 3 = B \Rightarrow B = 3 s = 0 : 0 = 3 A + B \Rightarrow A = - 1$

Therefore

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

Using the inverse Laplace transform we obtain the solution of given differential equation

$y (t) = L^{- 1} \{- \frac{1}{s + 3} + \frac{3}{{(s + 3)}^{2}}\} (t) = - L \{\frac{1}{s + 3}\} + 3 L \{\frac{1}{{(s + 3)}^{2}}\} = - e^{- 3 t} + 3 t e^{- 3 t}$

Therefore,

$y (t) = - e^{- 3 t} + 3 t e^{- 3 t}$

Therefore, the initial value for $y'' + 6 y' + 9 y = 0$ is $y (t) = - e^{- 3 t} + 3 t e^{- 3 t}$

Most popular questions for Math Textbooks

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

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

In Problems $1 - 14$ , solve the given initial value problem using the method of Laplace transforms.

$y'' - y = t - 2; y (2) = 3, y' (2) = 0$

Determine the inverse Laplace transform of the given function.

$\frac{6}{{(s - 1)}^{4}}$ .

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.]

$\cos^{3} t$

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

$e^{2 t} - t^{3} + t^{2} - s i n 5 t$

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