Mathematics 304, section 1 – Ordinary Differential Equations Exam 2 Solutions November 22, 2004

I. A) With a = 10, we have A = ( \begin{matrix}−1 & 10 \\ −1 & 0 \end{matrix} ) which has trace T = −1 and determinant

D = 10. Since D > T^{2}/4, we are in the region of the trace-determinant plane corresponding to systems with spiral sinks at the origin.

B) With a = −2, we have T = −1, D = −2. The characteristic polynomial is λ^{2} + λ − 2 = (λ + 2)(λ − 1) so the eigenvalues are λ = −2, 1. For λ = −2,

ker(A + 2I) = ker ( \begin{matrix}1 & −2 \\ −1 & 2 \end{matrix} ) = span ( \begin{matrix}2 \\ 1 \end{matrix} ) .

For λ = 1,

ker(A − I) = ker ( \begin{matrix}−2 & −2 \\ −1 & −1 \end{matrix} ) = span ( \begin{matrix}−1 \\ 1 \end{matrix} ) .

So the general solution looks like

X = c_{1}e^{−2t} ( \begin{matrix}2 \\ 1 \end{matrix} ) + c_{2}e^{t} ( \begin{matrix}−1 \\ 1 \end{matrix} ) ,

where c_{1} and c_{2} are arbitrary constants. Since the signs of the eigenvalues are opposite, the origin is a saddle point for this system.

C) As a varies over the real numbers, the motion in the trace-determinant plane takes us “up” the vertical line with T = −1. The systems in the family have saddle points if a < 0 (like a = −2 in part B), a line of equilibria if a = 0, an ordinary sink if 0 < a < 1/4, a sink from a repeated real root if a = 1/4, and spiral sinks for all a > 1/4 (like a = 10 in part A).

II. A) By the usual linearity of differentiation, We have

m(x_{p,1} + x_{p,2})′′ + b(x_{p,1} + x_{p,2})′ + k(x_{p,1} + x_{p,2})
= mx′′_{p,1} + mx′′_{p,2} + bx′_{p,1} + bx′_{p,2} + kx_{p,1} + kx_{p,2}
= (mx′′_{p,1} + bx′_{p,1} + kx_{p,1}) + (mx′′_{p,2} + bx′_{p,2} + kx_{p,2})
= g_{1}(t) + g_{2}(t)

Hence x = x_{p,1}(t) + x_{p,2}(t) is a particular solution of mx′′ + bx′ + kx = g_{1}(t) + g_{2}(t).

B) The characteristic polynomial of x′′ + 4x′ + 3x = 0 is λ^{2} + 4λ + 3 = (λ + 1)(λ + 3) so λ = −1, −3. The general solution of the homogeneous equation is

x_{h} = c_{1}e^{−t} + c_{2}e^{−3t}

where c_{i} are arbitrary constants. Next we find particular solutions for g_{1}(t) = e^{7t} and g_{2}(t) = 4e^{−t}. For g_{1}(t) (not a solution of the homogeneous equation), we take x_{p,1}(t) = Ae^{7t}. Then

x′′_{p,1} + 4x′_{p,1} + 3x_{p,1} = (49 + 28 + 3)Ae^{7t} = 80Ae^{7t} = e^{7t}

implies A = 1/80. For g_{2}(t) (which is a solution of the homogeneous equation), we take x_{p,2}(t) = Bte^{−t}. Then

x′′_{p,2} + 4x′_{p,2} + 3x_{p,2} = 0 · te^{−t} + 2Be^{−t} = 4e^{−t}

implies B = 2. By part A, our general solution looks like

x = x_{h} + x_{p,1} + x_{p,2} = c_{1}e^{−t} + c_{2}e^{−3t} + \frac{1}{80}e^{7t} + 2te^{−t}

The final step is to solve for c_{1}, c_{2} from the initial conditions. We get

\begin{aligned}x(0) = 0 & = c_{1} + c_{2} + \frac{1}{80} \\ x′(0) = 1 & = −c_{1} − 3c_{2} + \frac{7}{80} + 2 \end{aligned}

Solving simultaneously,

c_{1} = \frac{−45}{80} = \frac{−9}{16}, c_{2} = \frac{11}{20}.

III. Resonance occurs in solutions of an undamped, sinusoidally forced harmonic oscillator equation when the frequency of the forcing term is the same as the natural frequency of the solutions of the homogeneous equation. For 4x′′ + 400x = 0, we have 4λ^{2} + 400 = 0, which implies λ = ±10i. The natural frequency is 10, so resonance occurs for ω = 10. The reason this happens is that the particular solution of the equation with the forcing term is

x_{p} = At cos(ωt)

for some A. The extra factor of t means that the solutions are not bounded as t → +∞.

IV. For matrices as described in the problem, there are one block for λ = 1, and 2 blocks for λ = 2. So there are exactly two possible canonical forms:

( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix} ) , ( \begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{matrix} )

V. The equilibrium points of the system are found by setting the right sides equal to zero and solving for x, y. From −y(1 − y) = 0, we get y = 0 or y = 1. Similarly, x = 0 or x = −1. All combinations of an x and y value are possible, so there are four equilibria in all:

(0, 0), (0, 1), (−1, 0), (−1, 1).

Extra Credit

The given graph shows a system with centers at (0, 0), (−1, 1) and saddle points at (0, 1), (−1, 0). This cannot be the system (2) because the linearization of (2) at the origin is

X′ = ( \begin{matrix}0 & −1 \\ −2 & 0 \end{matrix} ) X.

This linear system has a saddle point at the origin (since the determinant is negative). Moreover, our general theorem says this is one case where the nonlinear system has the same type of equilibrium as the linearization. Hence (2) also has a saddle point at (0, 0).