Question

In problems 21-24 verify that the vector X_pgiven is a particular solution of the given non-homogeneous linear system.

$$\begin{gathered} \mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{1}& \mathbf{2}& \mathbf{3}\\ \mathbf{-}\mathbf{4}& \mathbf{2}& \mathbf{0}\\ \mathbf{-}\mathbf{6}& \mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathit{X}\mathbf{+}\mathbf{\left(}\begin{array}{c}\mathbf{-}\mathbf{1}\\ \mathbf{4}\\ \mathbf{3}\end{array}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{t} \\ {\mathit{X}}_{\mathit{p}}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{sin}\mathbf{3}\mathbf{t}\\ \mathbf{0}\\ \mathbf{cos}\mathbf{3}\mathbf{t}\end{array}\mathbf{\right)} \end{gathered}$$

Short answer

Yes, the given vector $X_p=\left(\begin{array}{c}\sin 3\mathrm{t}\\ 0\\ \cos 3\mathrm{t}\end{array}\right)$is a particular solution of the givennon-homogeneous linear system.

Step-by-step solution

Definition of general solution - Non-Homogeneous Systems

Let X_pbe a given solution of the non-homogeneous system on an interval, I ,then

${\mathit{X}}_{\mathbf{c}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathbf{X}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathbf{X}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathbf{X}}_{\mathbf{n}}$

Denote the general solution on the same interval of the homogeneous system.

Then, the general solution of the non-homogeneous system on the interval is given by,
$\mathit{X}={\mathit{X}}_c+{\mathit{X}}_p$

We are given,

$X\text{'}=\left(\begin{array}{ccc}1& 2& 3\\ -4& 2& 0\\ -6& 1& 0\end{array}\right)X+\left(\begin{array}{c}-1\\ 4\\ 3\end{array}\right)sin3t$ ...(1)

$X_p=\left(\begin{array}{c}\sin 3\mathrm{t}\\ 0\\ \cos 3\mathrm{t}\end{array}\right)$ ...(2)

Obtaining the derivative of given vector

Differentiate the given vector,

$X_p=\left(\begin{array}{c}\sin 3\mathrm{t}\\ 0\\ \cos 3\mathrm{t}\end{array}\right)$

role="math" localid="1663911189477" $X_p^{\text{'}}=\left(\begin{array}{c}3\cos 3\mathrm{t}\\ 0\\ -3\sin 3\mathrm{t}\end{array}\right)$ ...(3)

Substitute (2) and (3) in (1),

role="math" localid="1663911425263" $X_p^{\text{'}}=\left(\begin{array}{ccc}1& 2& 3\\ -4& 2& 0\\ -6& 1& 0\end{array}\right)X_p+\left(\begin{array}{c}-1\\ 4\\ 3\end{array}\right)sin3t$ ...(4)

role="math" localid="1663911285442" $\left(\begin{array}{c}3\cos 3\mathrm{t}\\ 0\\ -3\sin 3\mathrm{t}\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ -4& 2& 0\\ -6& 1& 0\end{array}\right)X_p+\left(\begin{array}{c}-1\\ 4\\ 3\end{array}\right)sin3t$

Multiply the matrices and Combine the like terms,

$$\begin{gathered} \left(\begin{array}{c}3\cos 3\mathrm{t}\\ 0\\ -3\sin 3\mathrm{t}\end{array}\right)=\left(\begin{array}{c}\sin 3t+3\cos 3t-\sin 3t\\ -4\sin 3t+4\sin 3t\\ -6\sin 3t+3\sin 3t\end{array}\right) \\ \left(\begin{array}{c}3\cos 3\mathrm{t}\\ 0\\ -3\sin 3\mathrm{t}\end{array}\right)=\left(\begin{array}{c}3\cos 3\mathrm{t}\\ 0\\ -3\sin 3\mathrm{t}\end{array}\right) \end{gathered}$$

i.e., we see that equation (4), $X_p^{\text{'}}=\left(\begin{array}{ccc}1& 2& 3\\ -4& 2& 0\\ -6& 1& 0\end{array}\right)X_p+\left(\begin{array}{c}-1\\ 4\\ 3\end{array}\right)sin3t$

Thus, the given vector $X_p=\left(\begin{array}{c}\sin 3\mathrm{t}\\ 0\\ \cos 3\mathrm{t}\end{array}\right)$is a particular solution of the givennon-homogeneous linear system.