Suppose I have a system of non-homogeneous linear first order differential equations:

$$ x'=A x+b_0+b_1t $$

Where $A$ is a $2\times2$ invertible matrix, $b_0$ and $b_1$ are:

$$ b_0 = \begin{pmatrix} r\\ u\\ \end{pmatrix}, \qquad b_1 = \begin{pmatrix} m\\ n\\ \end{pmatrix} $$

A particular solution for this system is given:

$$ x_p=g_1+g_2t $$

where $g_1$ and $g_2$ are constant vectors. Would I be able to find these constant vectors by determining them independently? Meaning I split them up like this:

$$ g_1=-A^{-1}b_0 \\ g_2=-A^{-1}b_1 $$

  • $\begingroup$ Why is this getting downvoted? $\endgroup$ – Steve Apr 12 '16 at 22:32

Never mind, I figured it out.

$$ A(g_1+g_2t)+b_0+b_1t $$

$$ =Ag_1+Ag_2t+b_0+b_1t $$

$$ =Ag_1+b_0+(Ag_2+b_1)t $$

$g_1$ and $g_2$ can now be determined by solving the above system of equations.


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