where \(A\) is the amplitude of the solution, \(\omega\) is the frequency of the solution, and \(\phi\) is the phase angle.
If we consider different forcing functions \(g(t)\) for the equation \begin x'' + px' + qx = g(t), \end functions that are periodic are especially important. Recall that a function \(g(t)\) is if \begin g(t + T) = g(t) \end for all \(t\) and some fixed constant \(T\text<.>\) The most familiar periodic functions are \begin g(t) = \sin \omega t \mbox < and >g(t) = \cos \omega t. \endThe for each of these two functions is \(2 \pi / \omega\) and the is \(\omega / 2 \pi\text<.>\) These two functions share the additional property that their average value is zero. That is,
\begin \frac \int_0^T g(t) \, dt = 0. \end We say that occurs in the differential equation \begin x'' + px' + qx = A \cos \omega t + B \sin \omega t. \endwe can use Euler’s formula, \(e^ = \cos \beta t + i \sin \beta t\) to derive a particular solution. That is, we will assume that our particular solution has the form
\begin x_c = x_\text + i x_\text \end and use the properties of complex numbers.In the next examples, we solve a differential equation with sinusoidal forcing in two ways: first, using the method of undetermined coefficients, and second using complexification.
To find a particular solution, we can use the method of undetermined coefficients and assume that the solution has the form
\begin x_p = A \cos 2t + B \sin 2 t. \end If we carry out the appropriate calculations, we will obtain a particular solution \begin x_p = - \frac \cos 2t + \frac \sin 2t. \end Thus, the general solution is \begin x = x_h + x_p = c_1 e^ + c_2 e^ - \frac \cos 2t + \frac \sin 2t. \end Notice that all solutions of (2.3.1) will approach the particular solution as \(t \to \infty\text\)Now let us solve \(x'' + 6 x' + 5x = \sin 2t\) using complex numbers. We begin by passing to a related complex differential equation
\begin x'' + 6x' + 5x = e^ = \cos 2t + i \sin 2t.\tag \endIf we assume that the equation above has a complex solution of the form \(x_c = x_\text
\begin \frac x_c+ 6 \frac x_c + 5 x_c & = \frac (x_\text + i x_\text) + 6 \frac (x_\text + i x_\text) + 5 (x_\text + i x_\text)\\ & = e^\\ & = \cos 2t + i \sin 2t. \end
Equating the real and imaginary parts of this equation, we obtain\begin x''_\text + 6 x'_\text + 5 x_\text & = \cos 2t\\ x''_\text + 6 x'_\text + 5 x_\text & = \sin 2t. \end
Thus, if we can find a complex solution \(x_Setting \((1+12i)Ae^<2it>\) equal to the desired right-hand side \(e^<2it>\) and solving for \(A\text\) we immediately see that
\begin A = \frac = \frac - \frac i. \end Therefore, the complex solution to (2.3.2) is\begin x_c & = A e^\\ & = \left( \frac - \frac i\right) \cdot (\cos 2t + i \sin 2t)\\ & = \left( \frac \cos 2t + \frac \sin 2t \right) + i \left( -\frac \cos 2t + \frac \sin 2t \right). \end
The imaginary part of this function is \begin x_\text = -\frac \cos 2t + \frac \sin 2t. \endBy our prior observations, \(x_>\) is a solution to \(x'' + 6x' + 5x = \sin 2t\text<.>\) (Note that this agrees with what we found in Example 2.3.1.)
Find (1) a particular solution and (2) a general solution for each of the following differential equations.
= A\cos(\omega t - \phi)\text\) where \(A\) is the amplitude of the solution, \(\omega\) is the frequency, and \(\phi\) is the phase angle.
\begin x_p(t) = \frac> \sin(2t + \theta) = \frac> \cos\left(2t + \theta - \frac<\pi> \right) = \frac >\cos\left(2t - \phi \right), \end
where \(\phi \approx 3.058451\text<.>\) We say that \(\phi\) is the of our solution. The amplitude of our solution is \(1/\sqrt\) and the period is \(\pi\) (Figure 2.3.4).
Find the complex solution particular solution, \(y_c\) to \(y'' + 10 y' + 34 y = e^<2it>\text<.>\) That is, find \(a\) for \(y_p = a e^<2it>\text<.>\)
Determine \(A\) and \(B\text<,>\) so that \(y_p = A \cos 2t + B \sin 2t\) is a particular solution to (2.3.3)
Plot the solution you found in Task 2.3.2.e, labeling the amplitude, period, and frequency of your solution.
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 (2.3.4), we find that
\begin (8 + 6i) A e^ = -2 e^. \end Thus, \(x_c\) is a solution if \begin A = \frac = - \frac + \frac i \end\begin x_c(t) & = \left(- \frac + \frac i \right) (\cos 3t + i \sin 3t)\\ & = \left( - \frac \cos 3t - \frac \sin 3t \right) + i \left( \frac \cos 3t - \frac \sin 3t \right). \end
The imaginary part of this function is the solution that we seek, \begin x_p = \frac \cos 3t - \frac \sin 3t. \end Thus, the general solution to (2.3.4) is \begin x(t) = c_1 e^ \cos 4t + c_2 e^ \sin 4t + \frac \cos 3t - \frac \sin 3t. \end Now suppose that \(x(0) = 0\) and \(x'(0) = 0\text<.>\) We can quickly determine that\begin x'(t) = c_1 e^(- \cos 4t - 4 \sin 4t) + c_2 e^( 4\cos 4t - \sin 4t) - \frac \cos 3t - \frac \sin 3t \end
To solve this initial value problem, we must solve the linear system \begin c_1 + \frac \amp = 0\\ - c_1 + 4 c_2 - \frac \amp = 0. \end We obtain \(c_1 = -3/25\) and \(c_2 = 9/100\text<,>\) and the solution to our initial value problem is\begin x(t) & = -\frac e^(- \cos 4t - 4 \sin 4t) + \frac e^( 4\cos 4t - \sin 4t) - \frac \cos 3t - \frac \sin 3t\\ & = -\frac (4 \cos 4t - 3 \sin 4t)) e^ - \frac \cos 3t - \frac \sin 3t \end
The graph of our solution is given in Figure 2.3.6.Since \(y = x'(t)\text<,>\) we can now graph the solution curve in the phase plane (Figure 2.3.7). 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.
On the other hand, suppose we have initial conditions \(x(0) = 2\) and \(x'(0) = 2\text<,>\) we can solve the linear system
\begin c_1 + \frac \amp = 2\\ -4 c_1 - c_2 - \frac \amp = 2. \end to obtain \(c_1 = 47/25\) and \(c_2 = 109/100\text<.>\) Thus, solution to our initial value problem is \begin x(t) = \frac e^ \cos 4t + \frac e^ \sin 4t + \frac \cos 2t - \frac \sin 2t. \end The graph of our solution is given in Figure 2.3.8.If we examine the phase plane for this solution (Figure 2.3.9), 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 \(\omega / 2 \pi\text<.>\) These average value of each of these functions is zero.
We can use Euler’s formula and complexification to solve the equation \begin x'' + px' + qx = g(t), \endwhere 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 x(t) = A \cos(\omega t - \phi), \endwhere \(A\) is the amplitude of the solution, \(\omega / 2 \pi\) is the frequency of the solution, and \(\phi\) is the phase angle.
Is it possible for solution curves to intersect in the phase plane of a nonautonomous system? Why or why not?