eNonHomoSystem

Wxample Find the particular solution to the follwing differential equation．

$\displaystyle {\bf X}^{\prime} = \left(\begin{array}{rr} 3&2\\ 1&2 \end{array}\right){\bf X} + \left(\begin{array}{c} e^{2t}\\ 2e^{2t} \end{array}\right)$

Answer $\det(A - \lambda I) = (\lambda - 4)(\lambda - 1) = 0$ . Then $\lambda = 4,1$ ．We find the eigen vector corresponding to $4$ . $(A - 4 I) = \left(\begin{array}{rr} -1&2\\ 1&-2 \end{array}\right)$ . Then the eigen vector is $\left(\begin{array}{c} 2\\ 1 \end{array}\right)$ ．Thus, ${\bf X}_{1} = \left(\begin{array}{c} 2\\ 1 \end{array}\right)e^{4t}$ is a solution. We next find the eigen vector corresonding to $1$ . $(A - I) = \left(\begin{array}{rr} 2&2\\ 1&1 \end{array}\right)$ . Then eigen vector is $\left(\begin{array}{c} 1\\ -1 \end{array}\right)$ . Thus, ${\bf X}_{2} = \left(\begin{array}{c} 1\\ -1 \end{array}\right)e^{t}$ is also a solution. From this, we find the fundamental matrix.

$\displaystyle \Phi(t) = \left(\begin{array}{rr} 2e^{4t}&e^{t}\\ e^{4t}&-e^{t} \end{array}\right)$

. To find the general solution, we solve the following system of linear equation.

$\displaystyle \Phi {\bf U}^{\prime} = {\bf F}$

$\displaystyle \left(\begin{array}{rr} 2e^{4t}&e^{t}\\ e^{4t}&-e^{t} \end{array... ...nd{array}\right) = \left(\begin{array}{c} e^{2t}\\ 2e^{2t} \end{array}\right)$

To solve this, we use Cramer's rule.

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{rr} e^{2t}&e^{t}\\... ...\ e^{4t}&-e^{t} \end{array}\right\vert} = \frac{-3e^{3t}}{-3e^{5t}} = e^{-2t}$

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{rr} 2e^{4t}&e^{2t}... ...&2e^{2t} \end{array}\right\vert}{-3e^{5t}} = \frac{3e^{6t}}{-3e^{5t}} = -e^{t}$

From this, we integrate with respect to $t$

to get

$\displaystyle u_{1} = -\frac{1}{2}e^{-2t}, \ u_{2} = -e^{t}$

Therefore, the particular solution is given by

$\displaystyle {\bf X_{p}}$	$\displaystyle =$	$\displaystyle \Phi {\bf U} = \left(\begin{array}{rr} 2e^{4t}&e^{t}\\ e^{4t}&-e... ...}\right)\left(\begin{array}{c} -\frac{1}{2}e^{-2t}\\ -e^{t} \end{array}\right)$
	$\displaystyle =$	$\displaystyle \begin{pmatrix}- 2e^{2t}\\ \frac{1}{2}e^{2t} \end{pmatrix}$

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