91Ó°ÊÓ

The given differential equation is:

$y^4-3y\text{'}\text{'}-8y=\sin t               ...\left(\mathbf{1}\right)$

Write the homogeneous differential equation of equation (1),

$y^4-3y\text{'}\text{'}-8y=0$

The auxiliary equation for the above equation,

$m^4-3m^2-8=0$

Solve the auxiliary equation,

$m^4-3m^2-8=0$

Let $m^2=t$,

$$\begin{gathered} t^2-3t-8=0 \\ t=\frac{3Â\pm \sqrt{9+32}}{2} \\ t=\frac{3Â\pm \sqrt{41}}{2} \end{gathered}$$

$m={\left(\frac{3Â\pm \sqrt{41}}{2}\right)}^2$

According to the method of undetermined coefficients, consider the particular solution is,

$y_p\left(t\right)=A\sin t+B\cos t                  ...\left(\mathbf{2}\right)$

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

$$\begin{gathered} y_p\text{'}\left(t\right)=A\cos t-B\sin t \\ {y}_p\text{'}\text{'}\left(t\right)=-A\sin t-B\cos t \\ {y}_p\text{'}\text{'}\text{'}\left(t\right)=-A\cos t+B\sin t \\ {y_p}^4\left(t\right)=A\sin t+B\cos t \end{gathered}$$

Substitute value of$y_p\left(t\right),  y_p\text{'}\text{'}\left(t\right)$ and${y_p}^4\left(t\right)$ in the equation (1),

$$\begin{gathered} y^4-3y\text{'}\text{'}-8y=\sin t \\ A\sin t+B\cos t-3\left(-A\sin t-B\cos t\right)-8\left(A\sin t+B\cos t\right)=\sin t \\ -4A\sin t-4B\cos t=\sin t \end{gathered}$$

Comparing the coefficients of the above equation;

$$\begin{gathered} -4A=1 \\ A=-\frac{1}{4} \\ B=0 \end{gathered}$$

Substitute values A and B in equation (2).

Therefore, the particular solution of the equation (1),

$$\begin{gathered} y_p\left(t\right)=A\sin t+B\cos t \\ {y}_p\left(t\right)=-\frac{1}{4}\sin t+\left(0\right)\cos t \\ {y}_p\left(t\right)=-\frac{1}{4}\sin t \end{gathered}$$