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Determine the general solution to the system $x'=Ax+g(t)$ for the given matrix $A$ and vector $g(t)$ below.

$\mathbf{A} = \left( \begin{array}{ccc} 2 & -5 \\ 1 & -2 \end{array} \right)$ , $\mathbf{g(t)} = \left( \begin{array}{ccc} -\cos(t) \\ \sin(t) \end{array} \right)$

Attempt: if $\mathbf{x(t)} = \left( \begin{array}{ccc} x_1(t) \\ x_2(t) \end{array} \right)$ and $\mathbf{x'(t)} = \left( \begin{array}{ccc} x_1'(t) \\ x_2'(t) \end{array} \right)$ then we have

$x_1'(t)=2x_1(t)-5x_2(t)-\cos t$

$x_2'(t)=x_1(t)-2x_2(t)+\sin t$

How do we solve this system? Thanks!

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  • $\begingroup$ do you know how to find two solutions of the homogeneous system $x' = Ax?$ $\endgroup$ – abel Jun 9 '15 at 14:53
  • $\begingroup$ I do not remember but I found how the method works from my book. $\endgroup$ – Ergin Suer Jun 9 '15 at 14:56
  • $\begingroup$ You can diagonalise and work in the complex domain to solve the problem. The matrix above has eigenvectors $(5 , 2\pm i)^T$. This lets you compute $e^{At}$. $\endgroup$ – copper.hat Jun 9 '15 at 15:13
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I'm assuming you know how to solve the homogeneous system $$x_h' = Ax_h$$ as $$x_h=e^{At}$$

HINT: Use eigenvalues/eigenvectors. Once you have the answer to the homogeneous solution, you can compute the final solution by noting that $$x = x_h+x_p$$ where $x_p$ is a particular solution to this nonhomogeneous system. Guessing $$x_p = \left(\begin{array}{ccc} c_1\cos{t}+c_2\sin{t}\\c_3\cos{t}+c_4\sin{t}\end{array}\right)$$ and plugging into the original system $x_p'=Ax_p+g$ will allow us to solve for these coefficients and therefore the particular solution.

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  • $\begingroup$ How do you guess the matrix $x_p$? Do you recommend some book about this? $\endgroup$ – Ergin Suer Jun 9 '15 at 16:14
  • $\begingroup$ You guess the particular solution vector (note: NOT a matrix) by looking at the forcing vector (or vector in your original system that is causing it to not be homogeneous). In this case, your $g(t)$ is a vector of cosines and sines, so you guess that your particular solution will be too. You probably already did this for single differential equations before learning about matrices. $\endgroup$ – NoseKnowsAll Jun 9 '15 at 16:23
  • $\begingroup$ @ErginSuer I often find that Paul's Online Notes are quite helpful whenever you're working on anything Calc I-III, Differential Equations, or Linear Algebra. tutorial.math.lamar.edu/Classes/DE/DE.aspx $\endgroup$ – NoseKnowsAll Jun 9 '15 at 16:24
  • $\begingroup$ I couldn't find the coefficients $c_1, c_2, c_3$ and $c_4$ by using the two equations. $\endgroup$ – Ergin Suer Jun 9 '15 at 17:11
  • $\begingroup$ You can split up those two equations by recognizing that the left and right hands of each equation have to be identically equal. Thus, the coefficients of the sines must be equal and the coefficients of the cosines must be equal. Now you have 4 equations and 4 unknowns. $\endgroup$ – NoseKnowsAll Jun 9 '15 at 17:17

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