Hello,

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

$\displaystyle

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:

$\displaystyle

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

$

$\displaystyle

\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:

$\displaystyle

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:

$\displaystyle

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

$

So now sub this into the original Equation which will give:

$\displaystyle \frac{dx_p}{dt} = \mathbf Ax_p + \mathbf g(t)$

and collect terms to get:

$\displaystyle

\mathbf {Aa} = \mathbf{a}

$

$\displaystyle

\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?

Thanks in Advance