91Ó°ÊÓ

The differential equation is,

$x\text{'}\text{'}-x\text{'}-2x=e^{t}cost-t^2+co{s}^{3}t                                ...\left(1\right)$

Write the homogeneous differential equation of equation (1),

$x\text{'}\text{'}-x\text{'}-2x=0$

The auxiliary equation for the above equation,

$m^2-m-2=0$

Solve the auxiliary equation,

$$\begin{gathered} m^2-m-2=0 \\ {m}^2-2m+m-2=0 \\ m\left(m-2\right)+1\left(m-2\right)=0 \\ \left(m+1\right)\left(m-2\right)=0 \end{gathered}$$

The roots of the auxiliary equation are,

$m_1=-1,      m_2=2$

The complementary solution of the given equation is,

$x_c=c_1{e}^{-t}+c_2{e}^{2t}$

One knows that,

$$\begin{gathered} cos3t=4co{s}^{3}t-3cost \\ co{s}^{3}t=\frac{1}{4}cos3t+\frac{3}{4}cost \end{gathered}$$

From equation (1),

$$\begin{gathered} x\text{'}\text{'}-x\text{'}-2x=e^{t}cost-t^2+co{s}^{3}t \\ x\text{'}\text{'}-x\text{'}-2x=e^{t}cost-t^2+\frac{1}{4}cos3t+\frac{3}{4}cost                            ...\left(2\right) \end{gathered}$$

The particular solution of equation (1) will be the linear combination of terms$e^{t}cost,  -t^2,  \frac{1}{4}cos3t,  \frac{3}{4}cost$ and their independent derivatives.

Therefore, the particular solution of equation (1),

$x_p=\left(Acost+Bsint\right)e^t+C{t}^2+Dt+E+Fcos3t+Gsin3t+Hcost+Isint$