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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Given that ${{\mathbf{y}}}_{1}\left(\mathbf{t}\right){\mathbf{=}}\frac{1}{4}{\mathbf{sin}}{\mathbf{2}}{\mathbf{t}}$ is a solution to ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{cos}}{\mathbf{2}}{\mathbf{t}}$ and ${{\mathbf{y}}}_{2}\left(\mathbf{t}\right){\mathbf{=}}\frac{t}{4}{\mathbf{-}}\frac{1}{8}$ is a solution to role="math" localid="1654930126913" ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{t}}$, use the superposition principle to find solutions to the following differential equations: $\left(\mathbf{a}\right){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{t}}{\mathbf{+}}{\mathbf{cos}}{\mathbf{2}}{\mathbf{t}}$$\left(\mathbf{b}\right){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{2}}{\mathbf{t}}{\mathbf{-}}{\mathbf{3}}{\mathbf{cos}}{\mathbf{2}}{\mathbf{t}}$$\left(\mathbf{c}\right){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{11}}{\mathbf{t}}{\mathbf{-}}{\mathbf{12}}{\mathbf{cos}}{\mathbf{2}}{\mathbf{t}}$

1. $\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{4}\mathrm{sin}2\mathrm{t}+\frac{\mathrm{t}}{4}-\frac{1}{8}$
2. $\mathrm{y}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{2}-\frac{1}{4}-\frac{3}{4}\mathrm{sin}2\mathrm{t}$
3. $\mathrm{y}\left(\mathrm{t}\right)=\frac{11\mathrm{t}}{4}-\frac{11}{8}-3\mathrm{sin}2\mathrm{t}$
See the step by step solution

## Step 1: Write the given equation.

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

## Step 2: Use the superposition principle to find solutions.

One needs to find solutions to the following differential equation.

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=\mathrm{t}+\mathrm{cos}2\mathrm{t}$

According to the method of the superposition principle,

For any constants ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ the function

role="math" localid="1654930755300" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{y}}_{1}\left(\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{y}}_{1}\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\left(\frac{1}{4}\mathrm{sin}2\mathrm{t}\right)+{\mathrm{c}}_{2}\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ is a solution to the differential equation.

Write the $\mathrm{t}+\mathrm{cos}2\mathrm{t}$ as a linear combination of $\mathrm{cos}2\mathrm{t}$ and $\mathbf{t}$.

Thus, superposition is,

$⇒1\left(\mathrm{cos}2\mathrm{t}\right)+1\left(\mathrm{t}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_{1}=1\\ {\mathrm{c}}_{2}=1\end{array}$

Substitute the value of ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ in the equation (3),

Therefore, the solution of a differential equation,

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=\left(1\right)\left(\frac{1}{4}\mathrm{sin}2\mathrm{t}\right)+\left(1\right)\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{1}{4}\mathrm{sin}2\mathrm{t}+\frac{\mathrm{t}}{4}-\frac{1}{8}\end{array}$

## Step 3: Use the superposition principle to find solutions

To find solutions to the following differential equation;

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=2\mathrm{t}-3\mathrm{cos}2\mathrm{t}$

According to the method of the superposition principle,

For any constants ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ the function

role="math" localid="1654931342988" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{y}}_{1}\left(\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{y}}_{1}\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\left(\frac{1}{4}\mathrm{sin}2\mathrm{t}\right)+{\mathrm{c}}_{2}\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ is a solution to the differential equation.

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

Hence, superposition is,

$⇒-3\left(\mathrm{cos}2\mathrm{t}\right)+2\left(\mathrm{t}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_{1}=-3\\ {\mathrm{c}}_{2}=2\end{array}$

So, the solution of a differential equation,

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=\left(-3\right)\left(\frac{1}{4}\mathrm{sin}2\mathrm{t}\right)+\left(2\right)\right)\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{2}-\frac{1}{4}-\frac{3}{4}\mathrm{sin}2\mathrm{t}\end{array}$

## Step 4: Use the superposition principle to find solutions

We need to find solutions to the following differential equation.

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=11\mathrm{t}-12\mathrm{cos}2\mathrm{t}$

According to the method of the superposition principle, for any constants ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ the function

role="math" localid="1654931947432" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{y}}_{1}\left(\mathrm{t}\right)+{\mathrm{c}}_{2}{\mathrm{y}}_{1}\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\left(\frac{1}{4}\mathrm{sin}2\mathrm{t}\right)+{\mathrm{c}}_{2}\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ is a solution to the differential equation.

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

Thus, superposition is,

$⇒-12\left(\mathrm{cos}2\mathrm{t}\right)+11\left(\mathrm{t}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_{1}=-12\\ {\mathrm{c}}_{2}=11\end{array}$

Substitute the value of ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ in the equation, we get:

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

Thereafter, the solution of the differential equation,

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