In Problems 33, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sint}$

The given differential equation is,

$\mathrm{y}\text{'}\text{'}\text{'}-\mathrm{y}\text{'}\text{'}+\mathrm{y}=\mathrm{sint}\dots \left(1\right)$

Consider the particular solution is,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asint}$

Take the first, second, and third derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=\mathrm{Acost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asint}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Acost}\end{array}$

Substitute value of${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$and${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)$in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}-\mathrm{y}\text{'}\text{'}+\mathrm{y}=\mathrm{sint}\\ -\mathrm{Acost}-(-\mathrm{Asint})+\mathrm{Asint}=\mathrm{sint}\\ -\mathrm{Acost}+2\mathrm{Asint}=\mathrm{sint}\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}-\mathrm{Acost}=0\Rightarrow \mathrm{A}=0\\ 2\mathrm{Asint}=\mathrm{sint}\Rightarrow \mathrm{A}=\frac{1}{2}\end{array}$

Here, A has two different values.

Thus,${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asint}$ is not a correct form of the particular solution.

Now, again Consider the particular solution is,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asint}+\mathrm{Bcost}\dots \left(2\right)$

Take the first, second, and third derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=\mathrm{Acost}-\mathrm{Bsint}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asint}-\mathrm{Bcost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Acost}+\mathrm{Bsint}\end{array}$

Substitute value of${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$and${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)$in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}-\mathrm{y}\text{'}\text{'}+\mathrm{y}=\mathrm{sint}\\ -\mathrm{Acost}+\mathrm{Bsint}-(-\mathrm{Asint}-\mathrm{Bcost})+\mathrm{Asint}+\mathrm{Bcost}=\mathrm{sint}\\ (-\mathrm{A}+2\mathrm{B})\mathrm{cost}+(\mathrm{B}+2\mathrm{A})\mathrm{sint}=\mathrm{sint}\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}-\mathrm{A}+2\mathrm{B}=0......\left(3\right)\\ -2\mathrm{A}+4\mathrm{B}=0......\left(4\right)\end{array}$

Solve the above equation,

$\begin{array}{l}(-\mathrm{A}+2\mathrm{B}=0)\times 2\\ -2\mathrm{A}+4\mathrm{B}=0\\ 2\mathrm{A}+\mathrm{B}=1\\ \overline{\begin{array}{l}5\mathrm{B}=1\\ \mathrm{B}=\frac{1}{5}\end{array}}\end{array}$

Substitute value B in the equation (3),

Substitute values A and B in the equation (2), we get:

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Asint}+\mathrm{Bcost}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{2}{5}\mathrm{sint}+\frac{1}{5}\mathrm{Bcost}\end{array}$

Therefore, the particular solution is:

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{2}{5}\mathrm{sint}+\frac{1}{5}\mathrm{Bcost}$