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 torole="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:

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

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

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

Given that ${\mathrm{y}}_1\left(\mathrm{t}\right)=\frac{1}{4}\sin 2\mathrm{t}$ is a solution to $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=\cos 2\mathrm{t}$androle="math" localid="1654930549569" ${\mathrm{y}}_2\left(\mathrm{t}\right)=\frac{\mathrm{t}}{4}-\frac{1}{8}$ is a solution to$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+4\mathrm{y}=\mathrm{t}$.

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

role="math" localid="1654930755300" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$ 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,

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

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

role="math" localid="1654931342988" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$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,

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

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

role="math" localid="1654931947432" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\frac{1}{4}\sin 2\mathrm{t}\right)+{\mathrm{c}}_2\left(\frac{\mathrm{t}}{4}-\frac{1}{8}\right)$is a solution to the differential equation.

Write the role="math" localid="1654932676605" $11\mathrm{t}-12\cos 2\mathrm{t}$ 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:

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

Thereafter, the solution of the differential equation,

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