# General solution to system of equations and initial value problem

## Homework Statement

Find the general solution for the following systems of equations, a solution to the
initial value problem and plot the phase portrait.
--> this is in matrix form

x' =

1 2
0 3

all multiplied by x.

also, x(0) =

2
-1

## Homework Equations

Determinant, etc.

## The Attempt at a Solution

I first use the eigen values and get it to:

eigenvalue1 = 1; eigenvalue2= 3

however, when I start with 1, my matrix is pretty screwed up:

0 2
0 2

So when I'm trying to figure out the values for the eigenvector I am confused. I don't know what to put it as.

Any help would be appreciated. Thanks.

tiny-tim
Homework Helper
hi paperweight11!
I first use the eigen values and get it to:

eigenvalue1 = 1; eigenvalue2= 3

how did you get those?

(those would be the eigenvalues for 1,0,,0,3)

HallsofIvy
Homework Helper
They are also the eigenvalues for
$$\begin{bmatrix} 1 & 2 \\ 0 & 3\end{bmatrix}$$
since the characteristic equation is just
$$\left|\begin{array}{cc}1- \lambda & 2 \\ 0 & 3- \lambda\end{array}\right|= (1- \lambda)(3- \lambda)= 0$$
(The eigenvalues for any triangular matrix are just the values on the diagonal.)

The eigenvectors must satisfy
$$\begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}$$
so x+ 2y= x and y= y. x+ 2y= x reduces to y= 0 which gives <1, 0> as an eigenvector, and
$$\begin{bmatrix}1 & 2 \\ 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y\end{bmatrix}$$
so x+ 2y= 3x and 3y= 3y so that y= x which gives <1, 1> as an eigenvector.

( Your
0 2
0 2
is the same as 2y= 0 so the eigenvector is <1, 0>.)