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10E

Expert-verified

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
$10 . y'' - 4 y = 4 t - 8 e^{- 2 t}; y (0) = 0, y' (0) = 5$

The Initial value for $y'' - 4 y = 4 t - 8 e^{- 2 t} i s y (t) = - t - e^{- 2 t} + 2 t e^{- 2 t} + e^{2 t}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define 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 as follows:

$L \{y'' - 4 y\} = L \{4 t - 8 e^{- 2 t}\} L \{y''\} - 4 L \{y\} = \frac{4}{s^{2}} - \frac{8}{s + 2}$

Solve the Laplace transform as:

$[s^{2} Y (s) - s y (0) - y' (0)] - 4 Y (s) = \frac{- 8 s^{2} + 4 s + 8}{s^{2} (s + 2)} s^{2} Y (s) - 5 - 4 Y (s) = \frac{- 8 s^{2} + 4 s + 8}{s^{2} (s + 2)} s^{2} Y (s) - 4 Y (s) = \frac{- 8 s^{2} + 4 s + 8}{s^{2} (s + 2)} + 5 (s^{2} - 4) Y (s) = \frac{5 s^{3} + 2 s^{2} + 4 s + 8}{s^{2} (s + 2)}$

Solve further as:

$Y (s) = \frac{5 s^{3} + 2 s^{2} + 4 s + 8}{s^{2} {(s + 2)}^{2} (s - 2)}$

Using partial fractions solve as:

$\frac{5 s^{3} + 2 s^{2} + 4 s + 8}{s^{2} {(s + 2)}^{2} (s - 2)} = \frac{A}{s^{2}} + \frac{B}{s + 2} + \frac{C}{{(s + 2)}^{2}} + \frac{D}{s - 2}$

Simplify the partial fractions as:

$5 s^{3} + 2 s^{2} + 4 s + 8 = \{A {(s + 2)}^{2} (s - 2) + B s^{2} (s + 2) (s - 2) + C s^{2} (s - 2) + S - D s^{2} {(s + 2)}^{2}\}$

Using s=0,2,1,-1 , respectively, gives

$s = 0 : 8 = - 8 A \Rightarrow A = - 1 s = 2 : 64 = 64 D \Rightarrow D = 1 s = 1 : 19 = 18 - 3 B - C \Rightarrow 1 = - 3 B - C s = - 1 : 1 = 4 - 3 B - 3 C \Rightarrow 3 = 3 B + 3 C$

Find B and C from the system.

$\{\begin{cases} \begin{array}{l} 3 B - C = 1 \\ 3 B + 3 C = 3 \end{array} \end{cases} \Rightarrow \{\begin{cases} \begin{array}{l} C = 2 \\ B = - 1 \end{array} \end{cases}$

$T h e r e f o r e, Y (s) = - \frac{1}{s^{2}} - \frac{1}{s + 2} + \frac{2}{{(s + 2)}^{2}} + \frac{1}{s - 2} U \sin g t h e i n v e r s e L a p l a c e t r a n s f o r m w e o b t a i n t h e s o l u t i o n o f g i v e n d i f f e r e n t i a l e q u a t i o n y (t) = L^{- 1} \{- \frac{1}{s^{2}} - \frac{1}{s + 2} + \frac{2}{{(s + 2)}^{2}} + \frac{1}{s - 2}\} (t) = - L \{\frac{1}{s^{2}}\} - L \{\frac{1}{s + 2}\} + 2 L \{\frac{1}{{(s + 2)}^{2}}\} + L^{- 1} \{\frac{1}{s - 2}\} = - t - e^{- 2 t} + 2 t e^{- 2 t} + e^{2 t} T h e r e f o r e, t h e i n i t i a l v a l u e f o r y'' - 4 y = 4 t - 8 e^{- 2 t} i s y (t) = - t - e^{- 2 t} + 2 t e^{- 2 t} + e^{2 t}$

Most popular questions for Math Textbooks

Use the convolution theorem to find the inverse Laplace transform of the given function. $\frac{s}{(s - 1) (s + 2)} [H i n t : \frac{s}{s - 1} = 1 + \frac{1}{s - 1}]$

In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.

$y'' + 5 y' + 6 y = g (t) : y (0) = 0, y' (0) = 2, w h e r e g (t) = {\begin{cases} 0, 0 ⩽ t < 1, \\ t, 1 < t < 5, \\ 1, 5 < t \end{cases}$

Find the Laplace transform of $f (t) = \int_{0}^{t} (t - v) e^{3 v} d v$

In Problems 1-10, determine the inverse Laplace transform of the given function.

$\frac{s - 1}{2 s^{2} + s + 6}$

Determine the inverse Laplace transform of the given function.

$\frac{e^{- 2 s} (4 s + 2)}{(s - 1) (s + 2)}$

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