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.
$z'' + 5 z' - 6 z = 21 e^{t - 1}, z (1) = - 1, z' (1) = 9$

The Initial value for $z'' + 5 z' - 6 z = 21 e^{t - 1}$ is $z (t) = 3 (t - 1) e^{t - 1} - e^{- 6 (t - 1)}$

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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

Shift the initial conditions to $t = 0$ by defining a new function:

$y (t) = z (t + 1) y' (t) = z' (t + 1) y'' (t) = z'' (t + 1)$

Replace $t$ by $t + 1$ in the condition

$z'' (t) + 5 z' (t) - 6 z (t) = 21 e^{t - 1} \Rightarrow z'' (t + 1) + 5 z' (t + 1) - 6 z (t + 1) = 21 e^{(t + 1) - 1}$

$\Rightarrow z'' (t + 1) + 5 z' (t + 1) - 6 z (t + 1) = 21 e^{t}$

Substitute $y (t) = z (t + 1)$

Solve for initial condition.

$y (0) = z (0 + 1) = z (1) = - 1$

Solve for differentiation of initial condition.

$y' (0) = z' (0 + 1) = z' (1) = 9$

Simplify the equation as:

$y'' (t) + 5 y' (t) - 6 y (t) = 21 e^{t} y (0) = - 1 y' (0) = 9$

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

sing the properties listed below, take the Laplace transform of the equation

$L \{y'\} (s) = s L \{y\} (s) - y (0) L \{y''\} (s) = s^{2} L \{y\} (s) - s y (0) - y' (0)$

$L \{e^{a t}\} (s) = \frac{1}{s - a} L \{y''\} + 5 L \{y'\} - 6 L \{y\} = 21 L \{e^{t}\}$

Substitute the properties into the equation.

$[s^{2} Y - s y (0) - y' (0)] + 5 [s Y - y (0)] - 6 Y = \frac{21}{s - 1}$

Substitute the initial conditions:

$y (0) = - 1 a n d y' (0) = 9$

$(s^{2} Y + s - 9) + 5 (s Y + 1) - 6 Y = \frac{21}{s - 1}$

Distribute and simplify:

$\Rightarrow s^{2} Y + 5 s Y - 6 Y + s - 4 = \frac{21}{s - 1}$

Isolate the Y variable.

$Y (s^{2} + 5 s - 6) = \frac{21}{s - 1} - s + 4 \Rightarrow Y (s^{2} + 5 s - 6) = \frac{21 + (- s + 4) (s - 1)}{s - 1} \Rightarrow Y = \frac{- s^{2} + 5 s + 17}{(s - 1) (s^{2} + 5 s - 6)}$

Find the partial fraction expansion

Because $s - 1$ is a repeated factor of ${(s - 1)}^{2} (s + 6)$ ,we include $(s - 1)$ and ${(s - 1)}^{2}$

$Y = \frac{- s^{2} + 5 s + 17}{(s - 1) (s - 1) (s + 6)}$

$\Rightarrow \frac{- s^{2} + 5 s + 17}{{(s - 1)}^{2} (s + 6)} = \frac{A}{s - 1} + \frac{B}{{(s - 1)}^{2}} + \frac{C}{s + 6}$

Combine the fractions to equate the numerators.

$- s^{2} + 5 s + 17 = A (s - 1) (s + 6) + B (s + 6) + C {(s - 1)}^{2}$

Solve for variables by setting values of S

$s = 1 \Rightarrow 21 = A \cdot 0 + 7 B + C \cdot 0 = 7 B \Rightarrow B = 3 s = - 6 \Rightarrow - 49 = A \cdot 0 + 3 \cdot 0 + 49 C \Rightarrow C = - 1$

$s = 0 \Rightarrow 17 = - 6 A + 3 \cdot 6 - 1 \cdot 1 \Rightarrow A = 0$

Substitute the values A,B,C of into partial fraction expansion

Using the properties listed below take the inverse Laplace transform to obtain the solution $y (t)$

$c^{- 1} \{\frac{1}{s - a}\} = e^{a t} c^{- 1} \{\frac{n!}{{(s - a)}^{n + 1}}\} = e^{a t} t^{n}$

Solve for the solution of differential equation as:

$y (t) = L^{- 1} \{Y\} = 3 L^{- 1} \{\frac{1}{{(s - 1)}^{2}}\} - L^{- 1} \{\frac{1}{s + 6}\} = 3 t e^{t} - e^{- 6 t}$

Since , $y (t - 1) = z (t)$ , , replace $t$ by $t - 1$

$y (t - 1) = z (t) = 3 (t - 1) e^{t - 1} - e^{- 6 (t - 1)}$

Therefore, the Initial value for $z'' + 5 z' - 6 z = 21 e^{t - 1}$ is $z (t) = 3 (t - 1) e^{t - 1} - e^{- 6 (t - 1)}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In Problems $1 - 14$ , solve the given initial value problem using the method of Laplace transforms.
$z'' + 5 z' - 6 z = 21 e^{t - 1}, z (1) = - 1, z' (1) = 9$

Step by Step Solution

Step 1: Define the Laplace Transform

Step 2: Determine the initial value of Laplace transform

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 1-14 , solve the given initial value problem using the method of Laplace transforms. z''+5z'-6z=21et-1, z1=-1, z'1=9

Step by Step Solution

Step 1: Define the Laplace Transform

Step 2: Determine the initial value of Laplace transform

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In Problems $1 - 14$ , solve the given initial value problem using the method of Laplace transforms.
$z'' + 5 z' - 6 z = 21 e^{t - 1}, z (1) = - 1, z' (1) = 9$