Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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  $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{t}$, use the superposition principle to find solutions to the following differential equations:

$\left(\mathbf{a}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{t}$

$\left(\mathbf{b}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{cos}\mathbf{2}\mathbf{t}$

$\left(\mathbf{c}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{11}\mathbf{t}\mathbf{-}\mathbf{12}\mathbf{cos}\mathbf{2}\mathbf{t}$

Step-by-Step Solution

Answer

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

Step 1: Write the given equation.

Given that  ${\mathrm{y}}_1\left(\mathrm{t}\right)=\frac{1}{4}\sin 2\mathrm{t}$  is a solution to $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=\cos 2\mathrm{t}$ and  ${\mathrm{y}}_2\left(\mathrm{t}\right)=\frac{\mathrm{t}}{4}-\frac{1}{8}$ 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.

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

 

According to the method of the superposition principle,

 

For any constants ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ the function

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$  is a solution to the differential equation.

 

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

 Thus, superposition is,

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

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,

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

Step 3: Use the superposition principle to find solutions

To find solutions to the following differential equation;

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

 

According to the method of the superposition principle,

 

For any constants ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ the function

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ is a solution to the differential equation.

 

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

 

Hence, superposition is,

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

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,

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

Step 4: Use the superposition principle to find solutions

We need to find solutions to the following differential equation.

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

 

According to the method of the superposition principle, for any constants ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ the function

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ is a solution to the differential equation.

 

Write the $11\mathrm{t}-12\cos 2\mathrm{t}$  as a linear combination of $\cos 2\mathrm{t}$ and  $\mathbf{t}$.

 

Thus, superposition is,

 

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

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:

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

Thereafter, the solution of the differential equation,

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