Q33E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. Do not solve.

$\mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{x}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{-}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{t}$

Step-by-Step Solution

Answer

Thus, the answer is:

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

Step 1: Firstly, write the auxiliary equation of the given differential equation

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,

$m^2-m-2=0$ 

Step 2: Now find the complementary solution of the given equation is

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,

$x_c=c_1{e}^{-t}+c_2{e}^{2t}$ 

Step 3: Now find the form of a particular solution

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),

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