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

Given that $y_{1} (t) = \frac{1}{4} \sin 2 t$ is a solution to $y'' + 2 y' + 4 y = \cos 2 t$ and $y_{2} (t) = \frac{t}{4} - \frac{1}{8}$ is a solution to role="math" localid="1654930126913" $y'' + 2 y' + 4 y = t$ , use the superposition principle to find solutions to the following differential equations:
$(a) y'' + 2 y' + 4 y = t + \cos 2 t$
$(b) y'' + 2 y' + 4 y = 2 t - 3 \cos 2 t$
$(c) y'' + 2 y' + 4 y = 11 t - 12 \cos 2 t$

$y (t) = \frac{1}{4} \sin 2 t + \frac{t}{4} - \frac{1}{8}$
$y (t) = \frac{t}{2} - \frac{1}{4} - \frac{3}{4} \sin 2 t$
$y (t) = \frac{11 t}{4} - \frac{11}{8} - 3 \sin 2 t$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given equation.

Given that $y_{1} (t) = \frac{1}{4} \sin 2 t$ is a solution to $y'' + 2 y' + 4 y = \cos 2 t$ and role="math" localid="1654930549569" $y_{2} (t) = \frac{t}{4} - \frac{1}{8}$ is a solution to $y'' + 2 y' + 4 y = t$ .

Step 2: Use the superposition principle to find solutions.

One needs to find solutions to the following differential equation.

$y'' + 2 y' + 4 y = t + \cos 2 t$

According to the method of the superposition principle,

For any constants $c_{1}$ and $c_{2}$ the function

role="math" localid="1654930755300" $y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (\frac{1}{4} \sin 2 t) + c_{2} (\frac{t}{4} - \frac{1}{8})$ is a solution to the differential equation.

Write the $t + \cos 2 t$ as a linear combination of $\cos 2 t$ and $t$ .

Thus, superposition is,

$\Rightarrow 1 (\cos 2 t) + 1 (t)$

The coefficients of the above equation are,

$\begin{matrix} c_{1} = 1 \\ c_{2} = 1 \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation (3),

Therefore, the solution of a differential equation,

$\begin{matrix} y (t) = (1) (\frac{1}{4} \sin 2 t) + (1) (\frac{t}{4} - \frac{1}{8}) \\ y (t) = \frac{1}{4} \sin 2 t + \frac{t}{4} - \frac{1}{8} \end{matrix}$

Step 3: Use the superposition principle to find solutions

To find solutions to the following differential equation;

$y'' + 2 y' + 4 y = 2 t - 3 \cos 2 t$

According to the method of the superposition principle,

For any constants $c_{1}$ and $c_{2}$ the function

role="math" localid="1654931342988" $y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (\frac{1}{4} \sin 2 t) + c_{2} (\frac{t}{4} - \frac{1}{8})$ is a solution to the differential equation.

Write the $2 t - 3 \cos 2 t$ as a linear combination of $\cos 2 t$ and $t$ .

Hence, superposition is,

$\Rightarrow - 3 (\cos 2 t) + 2 (t)$

The coefficients of the above equation are,

$\begin{matrix} c_{1} = - 3 \\ c_{2} = 2 \end{matrix}$

So, the solution of a differential equation,

$\begin{matrix} y (t) = (- 3) (\frac{1}{4} \sin 2 t) + (2)) (\frac{t}{4} - \frac{1}{8}) \\ y (t) = \frac{t}{2} - \frac{1}{4} - \frac{3}{4} \sin 2 t \end{matrix}$

Step 4: Use the superposition principle to find solutions

We need to find solutions to the following differential equation.

$y'' + 2 y' + 4 y = 11 t - 12 \cos 2 t$

According to the method of the superposition principle, for any constants $c_{1}$ and $c_{2}$ the function

role="math" localid="1654931947432" $y (t) = c_{1} y_{1} (t) + c_{2} y_{1} (t) \Rightarrow y (t) = c_{1} (\frac{1}{4} \sin 2 t) + c_{2} (\frac{t}{4} - \frac{1}{8})$ is a solution to the differential equation.

Write the role="math" localid="1654932676605" $11 t - 12 \cos 2 t$ as a linear combination of $\cos 2 t$ and $t$ .

Thus, superposition is,

$\Rightarrow - 12 (\cos 2 t) + 11 (t)$

The coefficients of the above equation are,

$\begin{matrix} c_{1} = - 12 \\ c_{2} = 11 \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation, we get:

role="math" localid="1654932914086" $\begin{matrix} y (t) = (- 12) (\frac{1}{4} \sin 2 t) + (11) (\frac{t}{4} - \frac{1}{8}) \\ y (t) = \frac{11 t}{4} - \frac{11}{8} - 3 \sin 2 t \end{matrix}$

Thereafter, the solution of the differential equation,

$y (t) = \frac{11 t}{4} - \frac{11}{8} - 3 \sin 2 t$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Write the given equation.

Step 2: Use the superposition principle to find solutions.

Step 3: Use the superposition principle to find solutions

Step 4: Use the superposition principle to find solutions

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Write the given equation.

Step 2: Use the superposition principle to find solutions.

Step 3: Use the superposition principle to find solutions

Step 4: Use the superposition principle to find solutions

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