Question

For each of the following ODEs, find the general solution, show the the general solution is fundamental, sketch a phase portrait, and classify the origin. 1 1. X. x= [1 +11 Now, finde, and ez for each problem given an initial condition (0) = [41]

Answer 1

We first need to find the eigenvalues of

$A = \begin{bmatrix} -\frac32 & 1 \\[5pt] - \frac 14 & -\frac 12 \end{bmatrix}$.

We calculate

$\det(A- \lambda I) = \begin{vmatrix} -\frac32-\lambda & 1 \\[5pt] - \frac 14 & -\frac 12-\lambda \end{vmatrix}$

-(3-1) (2 - 1)-(-+) X1

$= \left( \frac32 + \lambda\right ) \left( \frac 12 + \lambda \right )+ \frac14$
$= \frac 34 + \frac12\lambda + \frac 32 \lambda + \lambda^2 + \frac 14$
$= \lambda^2 + 2 \lambda + 1$
$= (\lambda + 1 )^2$

and so the (repeated) eigenvalue is $\lambda = -1$ . The first solution is given by

$\mathbf x_1(t) = e^{\lambda t} \mathbf v_1$

where $\mathbf v_1$ is an eigenvector of $\lambda = -1$ .

First, we have

$A- \lambda I = \begin{bmatrix} -\frac12& 1 \\[5pt] -\frac 1 4 & \frac12 \end{bmatrix}$

Since the second row is multiple of the first one, we only need to solve the equation $-\frac 12 x_1 + x_2 =0$ . Therefore, $\mathbf v_1 = \begin{bmatrix} 2x_2 \\ x_2\end{bmatrix}$ and with ,

22

$\mathbf x_1(t) = e^{-t} \begin{bmatrix} 2 \\1\end{bmatrix}$.

We now need to find the second solution given by
Xz(t) = vite + ve
where $\mathbf v_2$ satisfies $(A-\lambda I )\mathbf v_2 = \mathbf v_1$ . That is,

$\begin{bmatrix} -\frac12& 1 \\[5pt] -\frac 1 4 & \frac12 \end{bmatrix} \begin{bmatrix} v_1 \\[5pt] v_2 \end{bmatrix} = \begin{bmatrix} 2 \\[5pt] 1 \end{bmatrix}$.

The columns are not linearly independent, so we only consider the first one, $-\frac 12v_1 + v_2 = 2$ or . Taking we get that $\mathbf v_2 = \begin{bmatrix} -2 \\ 1\end{bmatrix}$ . The solution is then,

$\mathbf x_2(t) = e^{-t}t \begin{bmatrix} 2 \\1\end{bmatrix}+e^{-t} \begin{bmatrix} -2 \\1\end{bmatrix}$.

Here is the phase portrait

10 8 6 4 2 0 -2 -6 -8 -10 -10 -8 -6 -2 0 4 6 8 N 10

which shows that the origin is an unstable, improper node.

The general solution is given by

$\mathbf x(t) = c_1 \mathbf x_1 (t)+ c_2 \mathbf x_2 (t)$
$= c_1 e^{-t} \begin{bmatrix} 2 \\1\end{bmatrix}+ c_2 \left( e^{-t}t \begin{bmatrix} 2 \\1\end{bmatrix}+e^{-t} \begin{bmatrix} -2 \\1\end{bmatrix} \right )$ .

If $\mathbf x(0) = \begin{bmatrix} 1 \\ -1\end{bmatrix}$ then

$\begin{bmatrix} 1 \\ -1\end{bmatrix} = c_1 \begin{bmatrix} 2 \\1\end{bmatrix}+ c_2 \begin{bmatrix} -2 \\1\end{bmatrix} = \begin{bmatrix} 2 & -2\\1 & 1\end{bmatrix}\begin{bmatrix} c_1 \\c_2\end{bmatrix}$.

and therefore $c_1 = -\frac 14$ and $c_2 = -\frac 34$ .