An inhomogenous differential equation system is given as,
$$\left( \begin{matrix} \overset { . }{ x_{ 1 } } \\ \overset { . }{ x_{ 2 } } \end{matrix} \right) =\begin{pmatrix} -1 & 3 \\ -3 & -1 \end{pmatrix}\left( \begin{matrix} x_{ 1 } \\ x_{ 2 } \end{matrix} \right) +\left( \begin{matrix} 1 \\ 3 \end{matrix} \right) $$
Questions
i) Determine the complete complex solution to the corresponding homogeneous system.
(ii) Determine the complete real solution to the corresponding homogeneous system.
(iii) Determine the complete real solution to the given inhomogeneous system.
Approach I have solve the first two parts,
i) For the first part, $$x(t)=c_{ 1 }{ e }^{ (-1+3i) }\left( \begin{matrix} 1 \\ i \end{matrix} \right) +c_{ 2 }{ e }^{ (-1-3i) }\left( \begin{matrix} 1 \\ -i \end{matrix} \right)$$
ii) For the second part, I got the following, $$x(t)=c_{ 1 }{ e }^{ (-1t) }\left( \begin{matrix} cos(3t) \\ -sin(3t) \end{matrix} \right) +c_{ 2 }{ e }^{ (-1t) }\left( \begin{matrix} sin(3t) \\ cos(3t) \end{matrix} \right) $$
But I am stuck at the last part. I dont understand how to find the complete real solution for the nonhomogenous equation.
Any help would be great. Thank you.
There are two methods for this. Variation of parameters is one way, and it's basically a formula you just plug in for. The other is undetermined coefficients. You assume an solution of a certain form and then plug it in to the system and see what the constants have to be. In this case, assume your solution is $\left( \begin{matrix} x_{ 1 } \\ x_{ 2 } \end{matrix} \right)=\left( \begin{matrix} A \\ B \end{matrix} \right)$ where $A$ and $B$ are constants. Then
$$\left( \begin{matrix} 0 \\ 0 \end{matrix} \right) = \begin{pmatrix} -1 & 3 \\ -3 & -1 \end{pmatrix} \left( \begin{matrix} A \\ B \end{matrix} \right)+\left( \begin{matrix} 1 \\ 3 \end{matrix} \right)$$ From which you deduce that $A=2/5$ and $B=-1/5$. The final solution is your answer from part 2 added to the particular solution $\left( \begin{matrix} 2/5 \\ -1/5 \end{matrix} \right).$