# Second Order ODE with Matrix

Consider a (unforced) mass-spring system with mass $m = 1$, spring constant $k = 1$, and damping constant $c$, so that the displacement $x(t)$ satisfi es $x'' + cx' + x = 0$.

1. Re-write this ODE as a fir st-order system for x(t) = $\begin{bmatrix}x(t)\\y(t)\end{bmatrix}$ by setting $y=x'$.

So I did this and I have: $\begin{bmatrix}0 & 1\\-1 & -c\end{bmatrix}$

1. Find the general solution of this system and sketch the phase portrait for the three cases:

(a) $c = 0$ (undamped)

(b) $0 < c < 2$ (underdamped)

(c) $c > 2$ (overdamped)

So I'm having trouble with part 2. I think I can do (a) and I get complex roots for eigenvalues, but I don't understand how to do (b) and (c). Do I choose a fixed value for c? Like for (b) choose c = 1?

And I know I can check my solution just by comparing to solving the ODE not using matrices, but I don't really know how to solve for the general solution in question 2.

Thanks!

We can find the eigenvalues of the system as:

$$\lambda_{1,2} = \dfrac{1}{2}\left(-c \pm \sqrt{c^2 - 4} \right)$$

Now, we do a qualitative analysis for the different ranges of $c$ and see if we can determine the behavior of the eigenvalues in those ranges regardless of the value of $c$. We will then choose a representative $c$ if the behaviors show that the eigenvalues are the same within those ranges.

Case 1: c = 0

We have $\lambda_{1,2} = \pm ~ i$, which is a stable spiral. The phase portrait is:

Case 2: $0 \lt c \lt 2$

We have $\lambda_{1,2} =$ -real numer $\pm~$ imaginary number, which is a stable node. The phase portrait (sample $c = 1$) is:

Case 3: $c \gt 2$

We have $\lambda_{1,2} =$ both negative real, which is a stable node. The phase portrait (sample $c = 3$) is:

• Awesome! Three different cases, three very different phase portraits - You're in your element! +1 – Namaste Nov 18 '13 at 14:31
• This was reallllllllly!!! helpful. I didn't think of generalizing an equation for the eigenvalues of the system. Thanks so much. – Jay C Nov 18 '13 at 23:09
• Hi - is there any way you can explain a bit more about the general solution? I understand you have the eigenvalue, but do I then solve for the eigenvector? – Jay C Nov 19 '13 at 0:03
• You actually do not need the eigenvectors, although they are easy to find. For phase portraits, we only need the eigenvalues because it is a qualitative analysis. Regards – Amzoti Nov 19 '13 at 0:30
• Hi Amzoti - but the question is "find the general solution" and "sketch the phase portrait" so do I not need the eigenvectors to solve for the general solution? – Jay C Nov 20 '13 at 6:39