First-Order Linear Differential Problem

[Hiche]

Homework Statement

Solve the following IVP:

X' = \begin{pmatrix}2 & -1\\3 & -2\end{pmatrix}X + \begin{pmatrix}0\\t\end{pmatrix} with X(0) = \begin{pmatrix}1\\0\end{pmatrix}

Homework Equations

The Attempt at a Solution

The eigenvalue corresponding to \begin{pmatrix}2 & -1\\3 & -2\end{pmatrix} is \lambda = 0. We find that X_c = c_1\begin{pmatrix}1\\2\end{pmatrix} e^{0t}. Now in order to find X_p, how exactly is the right way? I took X_p = \begin{pmatrix}a_1\\b_1\end{pmatrix}t and wanted to find a_1 and b_1. Right or wrong?

 

[tiny-tim]

Hi Hiche!

The eigenvalue corresponding to \begin{pmatrix}2 & -1\\3 & -2\end{pmatrix} is \lambda = 0.

Nooo

 

[Hiche]

oh, crap! Apparently, 3 * 1 = 4 -_-

So, again, the eigenvalues are \lambda_1 = -1 and \lambda_2 = 1. I hope this is correct. So the solution of X_c = c_1\begin{pmatrix}1\\1\end{pmatrix}e^t + c_2\begin{pmatrix}1\\3\end{pmatrix}e^{-t}.

Now about X_p. Is my method correct (first post)?

 

[tiny-tim]

Now in order to find X_p, how exactly is the right way? I took X_p = \begin{pmatrix}a_1\\b_1\end{pmatrix}t and wanted to find a_1 and b_1. Right or wrong?

(i'm not sure, but…) i'd be inclined to go for X_p = \begin{pmatrix}a_0\\b_0\end{pmatrix} + \begin{pmatrix}a_1\\b_1\end{pmatrix}t

 

[Hiche]

Okay, so upon a little work, X_p = \begin{pmatrix}1\\2\end{pmatrix}t and the general solution is X = X_c + X_p

Thank you, tiny-tim.

 

[tiny-tim]

But X_p = \begin{pmatrix}1\\2\end{pmatrix}t isn't a solution …

X_p' = (1,2), and X_p(0) = (0,0)