functions that are periodic are especially important. Recall that a function \(g(t)\) is periodic if
\begin{equation*}
g(t + T) = g(t)
\end{equation*}
for all \(t\) and some fixed constant \(T\text{.}\) The most familiar periodic functions are
\begin{equation*}
g(t) = \sin \omega t \mbox{ and } g(t) = \cos \omega t.
\end{equation*}
The period for each of these two functions is \(2 \pi / \omega\) and the frequency is \(\omega / 2 \pi\text{.}\) These two functions share the additional property that their average value is zero. That is,
\begin{equation*}
a x'' + bx' + cx = A \cos \omega t + B \sin \omega t,
\end{equation*}
we can use Euler’s formula, \(e^{i \beta t} = \cos \beta t + i \sin \beta t\) to derive a particular solution. That is, we will assume that our particular solution has the form
\begin{equation*}
x_c = x_\text{Re} + i x_\text{Im}
\end{equation*}
and use the properties of complex numbers. 1
If complex numbers make you uncomfortable, the alternative is to become an expert in trigonometric identities
Now let us solve (4.3.1) using complex numbers. If we assume that the equation has a complex solution of the form \(x_c = x_\text{Re} + i x_\text{Im}\text{,}\) then
where \(\phi \approx 3.058451\text{.}\) We say that \(\phi\) is the phase angle of our solution. The amplitude of our solution is \(1/\sqrt{145}\) and the period is \(\pi\) (Figure 4.3.5).
This is a nonautonomous system, and the tangent vector of a solution curve in the phase plane depends not only on the position \((x, y)\text{,}\) but also on the time \(t\text{.}\) In other words, the direction field changes with time. Since the direction field changes with time, two solutions with the same \((x,y)\) value at different times can follow different paths. Consequently, solutions can cross each other in the \(xy\)-plane without violating the Existence and Uniqueness Theorem.
and we will use the Method of Undetermined Coefficients and assume that we can find a particular solution of the form \(x_c = A e^{3it}\text{.}\) Substituting \(x_c\) into equation (4.3.3), we find that
\begin{equation*}
(8 + 6i) A e^{3it} = -2 e^{3it}.
\end{equation*}
Thus, \(x_c\) is a solution if
\begin{equation*}
A = \frac{-2}{8 + 6i} = - \frac{4}{25} + \frac{3}{25} i
\end{equation*}
Since \(y = x'(t)\text{,}\) we can now graph the solution curve in the phase plane (Figure 4.3.9). Notice how the solution curve can intersect itself. The restoring force and damping are proportional to \(x\) and \(y = x'\text{,}\) respectively. When \(x\) and \(y\) are close to the origin, the external force is as large or larger than the restoring and damping forces. In this part of the \(xy\)-plane, the external force overcomes the damping and pushes the solution away from the origin.
If we examine the phase plane for this solution (Figure 4.3.11), we see that the initial damping and restoring forces are much larger than the external force. Thus, if we are far from the origin, the solutions in the \(xy\)-plane tend to spiral towards the origin and are similar to the solutions of the unforced equation.
The functions \(\sin \omega t\) and \(\cos \omega t\) are periodic with period \(2 \pi / \omega\) and frequency \(\omega / 2 \pi\text{.}\) The average value of each of these functions is zero.
where the forcing function \(g(t)\) is \(\sin \omega t\) or \(\cos \omega t\text{.}\) Furthermore, we can use complex numbers to express our solution in the form
\begin{equation*}
x(t) = A \cos(\omega t - \phi),
\end{equation*}
where \(A\) is the amplitude of the solution, \(\omega / 2 \pi\) is the frequency of the solution, and \(\phi\) is the phase angle.
we obtain a nonautonomous system. In this case the direction field changes with time, and two solutions with the same \((x,y)\) value at different times can follow different paths. Therefore, solutions can cross each other without violating the Existence and Uniqueness Theorem.
If we are far from the origin, the solutions in the \(xy\)-plane tend to spiral towards the origin and are similar to the solutions of the unforced equation. When \(x\) and \(y\) are close to the origin, the external force is as large or larger than the restoring and damping forces. In this part of the \(xy\)-plane, the external force overcomes the damping and pushes the solution away from the origin.
and taking the real part. Since \(2i\) is a root of the characteristic equation \(r^2 + 4 = 0\text{,}\) we seek a particular solution of the form \(y_p = ate^{2it}\text{.}\) Computing,
and taking the imaginary part. Since \(3i\) is not a root of the characteristic equation \(r^2 + 7r + 10 = 0\text{,}\) we try \(y_p = ae^{3it}\text{.}\) Substituting,
and taking the real part. Since \(2i\) is not a root of the characteristic equation \(r^2 + 2r + 2 = 0\) (whose roots are \(r = -1 \pm i\)), we try \(x_p = ae^{2it}\text{.}\) Substituting,
and taking the real part. Since \(i\) is not a root of the characteristic equation \(r^2 - 2r + 5 = 0\) (whose roots are \(r = 1 \pm 2i\)), we try \(x_p = ae^{it}\text{.}\) Substituting,
and taking the imaginary part. Since \(2i\) is not a root of the characteristic equation \(r^2 + 6r + 7 = 0\) (whose roots are \(r = -3 \pm \sqrt{2}\)), we try \(y_p = ae^{2it}\text{.}\) Substituting,
and taking the real part. Since \(2i\) is not a root of the characteristic equation \(r^2 + 4r + 13 = 0\) (whose roots are \(r = -2 \pm 3i\)), we try \(y_p = ae^{2it}\text{.}\) Substituting,
and taking the real part. Since \(3i\) is not a root of the characteristic equation \(r^2 + 6r + 8 = 0\) (whose roots are \(r = -2, -4\)), we try \(y_p = ae^{3it}\text{.}\) Substituting,
and taking the imaginary part. Since \(2i\) is not a root of the characteristic equation \(r^2 + 2r + 3 = 0\) (whose roots are \(r = -1 \pm i\sqrt{2}\)), we try \(x_p = ae^{2it}\text{.}\) Substituting,
and taking the imaginary part. Since \(3i\) is not a root of the characteristic equation \(r^2 + 4r + 20 = 0\) (whose roots are \(r = -2 \pm 4i\)), we try \(u_p = ae^{3it}\text{.}\) Substituting,
and taking the real part. Since \(5i\) is not a root of the characteristic equation \(r^2 + 4r + 20 = 0\) (whose roots are \(r = -2 \pm 4i\)), we try \(u_p = ae^{5it}\text{.}\) Substituting,
Finding Frequencies, Amplitudes, and Phase Angles.
Find a particular solution of the form \(y_p = A \cos(\omega t - \phi)\) for each equation in Exercise Group 4.3.5.11–17 and determine the frequency \(\omega\text{,}\) amplitude \(A\text{,}\) and phase angle \(\phi\) of the solution.
We complexify by solving \(y'' + 4y = 3e^{2it}\) and taking the real part. Since \(2i\) is a root of the characteristic equation \(r^2 + 4 = 0\text{,}\) this is a resonance case and we seek a particular solution of the form \(y_p = ate^{2it}\text{.}\) Computing,
Because the driving frequency \(\omega = 2\) equals the natural frequency of the system, the response is in resonance: the amplitude grows without bound as \(t \to \infty\) and the solution cannot be written in the steady-state form \(A\cos(\omega t - \phi)\text{.}\)
The natural frequency \(\omega_0 = 2\) equals the driving frequency, so this is a resonance case. The particular solution has the form \(y_p = t(A\cos 2t + B\sin 2t)\text{.}\) Substituting gives \(A = 0\) and \(B = \tfrac{3}{4}\text{,}\) so \(y_p = \tfrac{3t}{4}\sin 2t\text{.}\) The general solution is
Long-term behavior. Since the driving frequency exactly matches the natural frequency, the system is in resonance: the amplitude \(\tfrac{3t}{4}\) grows without bound as \(t \to \infty\text{.}\) The solution oscillates with increasing amplitude, and there is no steady state.
The characteristic equation \(r^2 + 7r + 10 = (r+2)(r+5) = 0\) has roots \(r = -2, -5\text{.}\) The particular solution from complexification is \(y_p = \tfrac{42\cos 3t - 2\sin 3t}{221}\text{,}\) so the general solution is
Long-term behavior. Since both roots are negative, the transient terms \(e^{-2t}\) and \(e^{-5t}\) decay to zero as \(t \to \infty\text{.}\) The solution approaches the steady-state periodic oscillation
\begin{gather*}
y \to \frac{42\cos 3t - 2\sin 3t}{221}
\quad \text{as } t \to \infty,
\end{gather*}
with frequency \(\omega = 3\) and amplitude \(\tfrac{2\sqrt{442}}{221} \approx 0.190\text{.}\)
The characteristic equation \(r^2 + 2r + 2 = 0\) has roots \(r = -1 \pm i\text{.}\) The particular solution from complexification is \(x_p = \tfrac{2\sin 2t - \cos 2t}{5}\text{,}\) so the general solution is
\begin{gather*}
x = e^{-t}(c_1\cos t + c_2\sin t)
+ \frac{2\sin 2t - \cos 2t}{5}.
\end{gather*}
Applying \(x(0) = 0\) gives \(c_1 = \tfrac{1}{5}\text{,}\) and \(x'(0) = 0\) gives \(-c_1 + c_2 + \tfrac{4}{5} = 0\text{,}\) so \(c_2 = -\tfrac{3}{5}\text{.}\) The solution is
\begin{gather*}
x = \frac{e^{-t}}{5}(\cos t - 3\sin t)
+ \frac{2\sin 2t - \cos 2t}{5}.
\end{gather*}
Long-term behavior. Since \(\text{Re}(r) = -1 \lt 0\text{,}\) the transient term \(e^{-t}(\cos t - 3\sin t)/5\) decays to zero as \(t \to \infty\text{.}\) The solution approaches the steady-state periodic oscillation
\begin{gather*}
x \to \frac{2\sin 2t - \cos 2t}{5}
\quad \text{as } t \to \infty,
\end{gather*}
with frequency \(\omega = 2\) and amplitude \(\tfrac{\sqrt{5}}{5} = \tfrac{1}{\sqrt{5}} \approx 0.447\text{.}\)
The characteristic equation \(r^2 - 2r + 5 = 0\) has roots \(r = 1 \pm 2i\text{.}\) The particular solution from complexification is \(x_p = \tfrac{6\cos t - 3\sin t}{10}\text{,}\) so the general solution is
\begin{gather*}
x = e^{t}(c_1\cos t + c_2\sin t)
+ \frac{6\cos t - 3\sin t}{10}.
\end{gather*}
Applying \(x(0) = 0\) gives \(c_1 + \tfrac{3}{5} = 0\text{,}\) so \(c_1 = -\tfrac{3}{5}\text{.}\) Differentiating and applying \(x'(0) = 0\text{:}\)
Long-term behavior. Since \(\text{Re}(r) = 1 \gt 0\text{,}\) the transient term \(e^t(-\tfrac{3}{5}\cos t +
\tfrac{9}{10}\sin t)\) grows without bound as \(t \to \infty\text{.}\) Unlike the damped case, there is no steady state: all solutions diverge and the system is unstable.
The characteristic equation \(r^2 + 6r + 7 = 0\) has roots \(r = -3 \pm \sqrt{2}\text{...}\) wait, discriminant=36−28=8, so \(r = -3 \pm \sqrt{2}\text{...}\) actually \(\sqrt{8}/2 = \sqrt{2}\text{,}\) giving \(r = -3 \pm \sqrt{2}\text{.}\) These are real roots, so the homogeneous solution is \(y_h = c_1 e^{(-3+\sqrt{2})t} + c_2 e^{(-3-\sqrt{2})t}\text{.}\) The particular solution from complexification is \(y_p = \tfrac{\sin 2t - 4\cos 2t}{17}\text{,}\) so the general solution is
So \(\sqrt{2}(c_1-c_2)=-\tfrac{2}{17}+\tfrac{12}{17}=\tfrac{10}{17}\text{,}\) giving \(c_1-c_2=\tfrac{10}{17\sqrt{2}}=\tfrac{5\sqrt{2}}{17}\text{.}\) Thus \(c_1=\tfrac{1}{17}(2+\tfrac{5\sqrt{2}}{2})\) and \(c_2=\tfrac{1}{17}(2-\tfrac{5\sqrt{2}}{2})\text{.}\) The solution is
Long-term behavior. Both roots \(-3 \pm \sqrt{2} \approx -1.586, -4.414\) are negative, so both transient terms decay to zero as \(t \to \infty\text{.}\) The solution approaches the steady-state oscillation
\begin{gather*}
y \to \frac{\sin 2t - 4\cos 2t}{17}
\quad \text{as } t \to \infty,
\end{gather*}
with frequency \(\omega = 2\) and amplitude \(\tfrac{1}{\sqrt{17}} \approx 0.243\text{.}\)
The characteristic equation \(r^2 + 4r + 13 = 0\) has roots \(r = -2 \pm 3i\text{,}\) giving the homogeneous solution \(y_h = e^{-2t}(c_1\cos 3t + c_2\sin 3t)\text{.}\) The particular solution from complexification is \(y_p = \tfrac{27\cos 2t + 24\sin 2t}{145}\text{,}\) so the general solution is
Long-term behavior. Since \(\text{Re}(r) = -2 \lt 0\text{,}\) the transient term decays to zero as \(t \to \infty\text{.}\) The solution approaches the steady-state periodic oscillation
\begin{gather*}
y \to \frac{27\cos 2t + 24\sin 2t}{145}
\quad \text{as } t \to \infty,
\end{gather*}
with frequency \(\omega = 2\) and amplitude \(\tfrac{3}{\sqrt{145}} \approx 0.249\text{.}\)
The characteristic equation \(r^2 + 6r + 8 = (r+2)(r+4) = 0\) has roots \(r = -2, -4\text{,}\) giving the homogeneous solution \(y_h = c_1 e^{-2t} + c_2 e^{-4t}\text{.}\) The particular solution from complexification is \(y_p = \tfrac{-\cos 3t + 18\sin 3t}{325}\text{,}\) so the general solution is
Long-term behavior. Since both roots are negative, the transient terms \(e^{-2t}\) and \(e^{-4t}\) decay to zero as \(t \to \infty\text{.}\) The solution approaches the steady-state periodic oscillation
\begin{gather*}
y \to \frac{-\cos 3t + 18\sin 3t}{325}
\quad \text{as } t \to \infty,
\end{gather*}
with frequency \(\omega = 3\) and amplitude \(\tfrac{1}{5\sqrt{13}} \approx 0.055\text{.}\)
