Question 3.7.2: Find the general solution of the nonhomogeneous system given...

We first find x_{h}. By finding the eigenvalues and eigenvectors of the coefficient matrix A, it is straightforward to show that

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

Therefore, the fundamental solution matrix is

Φ(t) =\begin{bmatrix} e^{−t}&e^{t}\\3e^{−t}&e^{t} \end{bmatrix}

Moreover, we can show that

Φ(t)^{-1}=-\frac {1}{2}  \begin{bmatrix} e^{t}&-e^{t}\\-3e^{−t}&e^{-t} \end{bmatrix}

We are now ready to compute x_{p} and write

x_{p}(t ) =Φ(t) \int {(t)^{-1} b(t ) dt}
=\begin{bmatrix} e^{−t}&e^{t}\\3e^{−t}&e^{t} \end{bmatrix} \int {-\frac {1}{2}  \begin{bmatrix} e^{t}&-e^{t}\\-3e^{−t}&e^{-t} \end{bmatrix} \begin{bmatrix} 1/(e^{t} +1)\\1 \end{bmatrix} dt }

=\begin{bmatrix} e^{−t}&e^{t}\\3e^{−t}&e^{t} \end{bmatrix} \int { \begin{bmatrix} \frac {1}{2} \frac {e^{2t}}{e^{t}+1}\\\frac {1}{2} \frac {2e^{-t}-1}{e^{t}+1}\end{bmatrix} dt}

At this point, it is easiest to use a computer algebra system to integrate and complete our calculation of x_{p}. Doing so, and then finding the required matrix product, we have

x_{p}(t )=\begin{bmatrix} e^{−t}&e^{t}\\3e^{−t}&e^{t} \end{bmatrix} \begin{bmatrix} \frac {1}{2}e^{t} − \frac {1}{2}ln(e^{t}+1)\\−e^{−t} − \frac {3}{2}t +\frac {3}{2} ln(e^{t}+1)\end{bmatrix}

=\begin{bmatrix} −\frac {1}{2}− \frac {1}{2}e^{−t} ln(e^{t}+1)−\frac {3}{2}te^{t} +\frac {3}{2} e^{t} ln(e^{t} +1)\\\frac {1}{2}− \frac {3}{2}e^{−t} ln(e^{t} +1)− \frac {3}{2}te^{t} + \frac {3}{2}e^{t} ln(e^{t} +1)\end{bmatrix}

Hence, the general solution to the given nonhomogeneous 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 {1}{2}− \frac {1}{2}e^{−t} ln(e^{t}+1)−\frac {3}{2}te^{t} +\frac {3}{2} e^{t} ln(e^{t} +1)\\\frac {1}{2}− \frac {3}{2}e^{−t} ln(e^{t} +1)− \frac {3}{2}te^{t} + \frac {3}{2}e^{t} ln(e^{t} +1)\end{bmatrix}