Consider the systems of differential equations
\[
\begin{array}{l}
\frac{d x}{d t}=0.5 x-0.4 y \\
\frac{d y}{d t}=-0.4 x+1.1 y .
\end{array}
\]

For this system, the smaller eigenvalue is 0.3 and the larger eigenvalue is 1.3 .

Use the phase plotter pplane9.m in MATLAB to determine how the solution curves behave.
A. The solution curves converge to different points.
B. The solution curves race towards zero and then veer away towards infinity. (Saddle)
C. All of the solution curves run away from 0 . (Unstable node)
D. All of the solution curves converge towards 0 . (Stable node)

The solution to the above differential equation with initial values $x(0)=5, y(0)=4$ is
\[
\begin{array}{l}
x(t)=\square \\
y(t)=\square
\end{array}
\]

Step 1 ：The given system of differential equations is: \[\begin{array}{l} \frac{d x}{d t}=0.5 x-0.4 y \\ \frac{d y}{d t}=-0.4 x+1.1 y . \end{array}\]

Step 2 ：The eigenvalues of the system are given as 0.3 and 1.3.

Step 3 ：The behavior of the solution curves can be determined by the eigenvalues. Since both eigenvalues are positive, the system is unstable. Therefore, all of the solution curves run away from 0. So, the correct answer is C. All of the solution curves run away from 0 (Unstable node).

Step 4 ：Now, let's solve the system with the initial values \(x(0)=5, y(0)=4\).

Step 5 ：The general solution of a system of differential equations can be written as: \[x(t) = c_1e^{\lambda_1t}v_1 + c_2e^{\lambda_2t}v_2\] \[y(t) = c_1e^{\lambda_1t}w_1 + c_2e^{\lambda_2t}w_2\] where \(\lambda_1\) and \(\lambda_2\) are the eigenvalues, \(v_1, v_2, w_1, w_2\) are the components of the corresponding eigenvectors, and \(c_1, c_2\) are constants determined by the initial conditions.

Step 6 ：To find the exact solution, we need to know the eigenvectors corresponding to the eigenvalues and solve for the constants \(c_1, c_2\). However, without this information, we cannot provide the exact solution for \(x(t)\) and \(y(t)\).