91Ó°ÊÓ

Given that ${\mathrm{y}}_1\left(\mathrm{t}\right)=\mathrm{cost}$is a solution to $\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+\mathrm{y}=\mathrm{sint}$and role="math" localid="1654927072726" ${\mathrm{y}}_2\left(\mathrm{t}\right)=\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}$is a solution to$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+\mathrm{y}={\mathrm{e}}^{2\mathrm{t}}.$

We need to find solutions to the following differential equation.

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+\mathrm{y}=5\mathrm{sint}$

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

role="math" localid="1654927221542" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\mathrm{cost}\right)+{\mathrm{c}}_2\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)$is a solution to the differential equation.

Write the$5\mathrm{sint}$as a linear combination of $\mathrm{sint}$ and ${\mathrm{e}}^{2\mathrm{t}}$.

Thus, superposition is,

$⇒5\left(\mathrm{sint}\right)+0\left({\mathrm{e}}^{2\mathrm{t}}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_1=5\\ {\mathrm{c}}_2=0\end{array}$

Substituting the value of ${\mathrm{c}}_1$and ${\mathrm{c}}_2$, we get:

role="math" localid="1654927468223" $\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=5\left(\mathrm{cost}\right)+0\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=5\mathrm{cost}\end{array}$

Therefore, the solution of a differential equation,

$\mathrm{y}\left(\mathrm{t}\right)=5\mathrm{cost}$

To solutions to the following differential equation;

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+\mathrm{y}=\mathrm{sint}-3{\mathrm{e}}^{2\mathrm{t}}$

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

role="math" localid="1654927831415" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\mathrm{cost}\right)+{\mathrm{c}}_2\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)$is a solution to the differential equation.

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

Thus, superposition is,

$⇒1\left(\mathrm{sint}\right)-3\left({\mathrm{e}}^{2\mathrm{t}}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_1=1\\ {\mathrm{c}}_2=-3\end{array}$

Substitute the value of ${\mathrm{c}}_1$and ${\mathrm{c}}_2$in the equation (3),

Hence, the solution of the differential equation,

role="math" localid="1654928094323" $\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=1\left(\mathrm{cost}\right)-3\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=\mathrm{cost}-{\mathrm{e}}^{2\mathrm{t}}\end{array}$

To find solutions to the following differential equation;

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+\mathrm{y}=4\mathrm{sint}+18{\mathrm{e}}^{2\mathrm{t}}$

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

role="math" localid="1654928569697" $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{y}}_1\left(\mathrm{t}\right)+{\mathrm{c}}_2{\mathrm{y}}_1\left(\mathrm{t}\right)⇒\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\left(\mathrm{cost}\right)+{\mathrm{c}}_2\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)$is a solution to the differential equation.

Write the $\mathbf{4}\mathbf{sint}\mathbf{+}\mathbf{18}{\mathbf{e}}^{2t}$as a linear combination of $\mathbf{sint}$and ${\mathbf{e}}^{2t}$

Thus, superposition is,

$⇒4\left(\mathrm{sint}\right)+18\left({\mathrm{e}}^{2\mathrm{t}}\right)$

The coefficients of the above equation are,

$\begin{array}{c}{\mathrm{c}}_1=4\\ {\mathrm{c}}_2=18\end{array}$

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

role="math" localid="1654928809131" $\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=4\left(\mathrm{cost}\right)+18\left(\frac{{\mathrm{e}}^{2\mathrm{t}}}{3}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=4\mathrm{cost}+6{\mathrm{e}}^{2\mathrm{t}}\end{array}$

Therefore, the solution of the differential equation,

$\mathrm{y}\left(\mathrm{t}\right)=4\mathrm{cost}+6{\mathrm{e}}^{2\mathrm{t}}$