0
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Why is this getting downvoted? $\endgroup$ – Steve Apr 12 '16 at 22:32
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.