Question 3.6.2: Find the general solution of the nonhomogeneous system given...

Since the eigenvalues of A =\begin{bmatrix}2 & -1 \\ 3 & -2 \end{bmatrix} are λ_{1} =−1 and λ_{2} = 1 with corresponding eigenvectors v_{1} = [1   3]^{T} and v_{2} = [1   1]^{T}, it follows that the complementary solution to the related homogeneous system is

xh = c_{1}e^{−t} \begin{bmatrix}1 \\ 3 \end{bmatrix}+c_{2}e^{t}\begin{bmatrix}1 \\ 1 \end{bmatrix}

To determine the particular solution x_{p} to the given nonhomogeneous system, we need to find a vector function x(t ) that simultaneously satisfies the system (3.6.9). Due to the presence of cos2t in the vector b, it is natural to guess that the components of x_{p} will somehow involve cos2t . In addition,since x_{p}' plays a role in the system, we must account for the possibility that the derivative of cos2t may also arise; moreover, since Ax will also be computed, linear combinations of vectors that involve the entries in x will be present. Therefore, we make the reasonable guess that x_{p} has the form

x_{p}=\begin{bmatrix}a cos2t +b sin2t \\ c cos2t +d sin2t\end{bmatrix}                  (3.6.10)

and attempt to determine values for the undetermined coefficients a,b, c, and d that make x_{p} a solution to the system.

We accomplish this by direct substitution into (3.6.9). First, observe that

\acute{x_{p}}=\begin{bmatrix}−2a sin2t +2b cos2t \\ −2c sin2t +2d cos2t\end{bmatrix}                  (3.6.11)

Now substituting (3.6.10) and (3.6.11) into (3.6.9), it follows

\begin{bmatrix}−2a sin2t +2b cos2t \\ −2c sin2t +2d cos2t\end{bmatrix} =\begin{bmatrix}2 & -1 \\ 3 & -2 \end{bmatrix} \begin{bmatrix}a cos2t +b sin2t \\ c cos2t +d sin2t\end{bmatrix}+\begin{bmatrix}2cos t \\ 0 \end{bmatrix}

If we now expand the matrix product and factor out the terms involving sin2t and cos2t on the right side,

−2a sin2t +2b cos2t = (2b −d) sin2t +(2a −c +1) cos2t            (3.6.12)
−2c sin2t +2d cos2t = (3b −2d) sin2t +(3a −2c) cos2t            (3.6.13)

In (3.6.12), we can equate the coefficients of sin2t to find that −2a = 2b −d. Doing likewise for the coefficients of cos2t, 2b = 2a −c +1. Similarly, (3.6.13) results in the two equations −2c = 3b − 2d and 2d = 3a − 2c. Reorganizing these four equations in four unknowns, we see that a,b, c, and d must satisfy the system

−2a −2b +d = 0
−2a +2b +c = 1
−3b −2c +2d = 0
−3a +2c +2d = 0

Row-reducing,

\begin{bmatrix}-2& -2&0&1&0 \\-2 & 2&1&0&1 \\ 0&-3&-2&2&0 \\ -3&0&2&2&0 \end{bmatrix}→\begin{bmatrix}1 & 0&0&0&-2/5 \\-0 & 1&0&0&2/5 \\ 0&0&1&0&-3/5 \\ 0&0&0&1&0 \end{bmatrix}

which shows a =−2/5, b = 2/5, c =−3/5, and d = 0, so a particular solution to the nonhomogeneous system is

x_{p} =\begin{bmatrix}−\frac{2}{5} cos2t + \frac{2}{5}sin2t\\ −\frac{3}{5} cos2t\end{bmatrix}

Finally, it follows that the general solution to the system is

x = x_{h} +x_{p} =c_{1}e^{−t} \begin{bmatrix}1 \\ 3 \end{bmatrix}+c_{2}e^{t}\begin{bmatrix}1 \\ 1 \end{bmatrix} + \begin{bmatrix}−\frac{2}{5} cos2t + \frac{2}{5}sin2t\\ −\frac{3}{5} cos2t\end{bmatrix}