# Thread: Method of undetermined coefficients - Nonhomogeneous Linear Systems

1. ## Method of undetermined coefficients - Nonhomogeneous Linear Systems

Hello,

I am simply trying to understand the way this system was solved in the text:

$
x^{\prime} = \left(\begin{array}{cc}2&-1\\3&-2\end{array}\right)x\ +\ \left(\begin{array}{c}e^t\\t\end{array}\right)
$

The two eigenvalues and corresponding eigenvectors that were found were:

$
\lambda_{1,2} = 1, -1
$

$
\xi^{(1)} = \left(\begin{array}{c}1\\1\end{array}\right)\ \xi^{(2)} = \left(\begin{array}{c}1\\3\end{array}\right)
$

Therefore we know the homogeneous part of the solution. Now Since:
$
g(t) = \left(\begin{array}{c}1\\0\end{array}\right)e^t + \left(\begin{array}{c}0\\1\end{array}\right)t
$

we guess by method of undetermined co-efficients that the particular solution has the form:

$
x_p(t) = \mathbf ate^t + \mathbf be^t + \mathbf ct + \mathbf d
$

So now sub this into the original Equation which will give:
$\frac{dx_p}{dt} = \mathbf Ax_p + \mathbf g(t)$
and collect terms to get:
$
\mathbf {Aa} = \mathbf{a}
$

$
\mathbf {Ab} = \mathbf a + \mathbf b - \left(\begin{array}{c}1\\0\end{array}\right)
$

etc.

I understand how to find a, c, d. But I'm not following the steps needed to find b, I know that we can find a because it is an eigenvalue of A, but with this info. how do we now find b?