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Lecture 2

Lecture 1 | Lecture 3

By Giuseppe Tenti and annotated by Douglas Wilhelm Harder.


In general, we will be discussing a single unknown function in a single variable. For example, we may consider the function tex:$$f(x)$$ and the variable tex:$$x$$; the function tex:$$q(t)$$ and the variable tex:$$t$$; or the function tex:$$y(t)$$ and the variable tex:$$t$$. It unnecessarily obfuscates the mathematics to include the variable in which the function is defined, and therefore we will use, for example, tex:$$y$$ and the variable is implied from the context.

For other functions, the variable will always be defined. For example, a constant voltage or resistance will be represented by a symbol, tex:$$v$$ or tex:$$R$$, but if the voltages and resistances are variable, tex:$$v(t)$$ or tex:$$R(t)$$, respectively.

We we define a function, we will also use indicate the variable, and therefore, the notation tex:$$v(t) = t\cos(t)$$ defines the function tex:$$v:t \rightarrow t\cos(t)$$.

Solving Differential Equations (MEM 10.4)

From the previous lecture, we saw that we may use differential equations to describe natural phenomena. For example, the circuit was described by the differential equation

tex:$$L\ddot{q} + R\dot{q} + \frac{1}{C} q = 0$$.

Let us consider, for example, an easier case where the differential equation is defined by a circuit with a voltage source tex:$$v(t)$$, a resistor with resistance tex:$$R$$, and a capacitor with capacitance tex:$$C$$, all in a single loop. This yields the differential equation:

tex:$$\dot{q} + \frac{1}{RC}q = v(t)$$.

If we assume that tex:$$RC = 1$$ and the voltage source is zero, we get the simplified differential equation

tex:$$\dot{q} + q = 0$$.

In this case, we notice that if we define tex:$$q(t) = e^{-t}$$, this is a function where the function plus the derivative is 0 and therefore, this function satisfies the differential equation. However, the functions tex:$$q(t) = 2e^{-t}$$ and tex:$$q(t) = 3e^{-t}$$ also satisfy the differential equation. In each case, the function gives us a curve which describes how the charge on the capacitor decreases as time grows. Figure 1 shows an entire collection of functions, all of which satisfy the differential equation: the value of the function is always equal to the negative of the slope.

Figure 1. Numerous solutions to the differential equation tex:$$\dot{q} + q = 0$$.

Suppose we change the voltage source to be a constant source of one volt. Now the differential equation is

tex:$$\dot{q} + q = 1$$.

and various solutions include tex:$$q(t) = 1$$ and tex:$$q(t) = 1 + e^{-t}$$, and Figure 2 shows an entire collection of such solutions:

Figure 2. Numerous solutions to the differential equation tex:$$\dot{q} + q = 1$$.

If we now change the voltage source to be an oscillating function tex:$$v(t) = \cos(t)$$, we get a different class of solutions, as is shown in Figure 3.

Figure 3. Numerous solutions to the differential equation tex:$$\dot{q} + q = \cos(t)$$.

Now, it appears that the charge on the capacitor begins to oscillate, too, but the oscillation appears to look more like tex:$$q(t) = \frac{1}{2}\cos(t) + \frac{1}{2}\sin(t)$$.

One important fact the reader may noted is that the solutions to these three 1st-order ordinary differential equations never cross. They may get arbitrarily close, but they do not appear to cross each other. This is, in fact, expected but is usually only proved in an upper-level course on differential equations.

We will conclude with two more examples. We will introduce a variable resistor and a switch.

Suppose we now add a variable resistor where the resistance is given by the function tex:$$R(t) = \frac{1}{1 + t}\ {\rm \Omega}$$ as time increases. The differential equation is now

tex:$$\dot{q} + \frac{1}{1 + t}q = 0$$.

Figure 4 shows numerous solutions to this differential equation.

Figure 4. Numerous solutions to the differential equation tex:$$\dot{q}(t) + \frac{1}{1 + t}q = 0$$.

The solutions decay to 0 like they did in Figure 1, but they do not decay as quickly: the greater resistance prevents the system from

Issue: The denominator of the tex:$$\frac{1}{RC}q$$ indicates that the behaviour would be identical whether we have a variable resistor with tex:$$R(t) = \frac{1}{1 + t}\ {\rm \Omega}$$ and tex:$$C = 1\ {\rm H}$$, or a variable capacitor with tex:$$R = 1\ {\rm \Omega}$$ and tex:$$C(t) = \frac{1}{1 + t}\ {\rm H}$$. Explain.

As an alternate example, suppose we have a voltage source which is initially 0 but after half a second, that is, at time tex:$$t = 0.5$$, the voltage is switch to two volts. Engineers use the unit step function, tex:$$u(t)$$, to represent such switches and it is defined as

tex:$$u(t) = \cases{
$0$ & $t < 0$\cr
$1$ & $t \ge 0$ }$$.

In this case, we may represent the turning on of the switch by the function tex:$$2u(t - 0.5)$$. Thus, the differential equation is now

tex:$$\dot{q} + q = 2u(t - 0.5)$$

and Figure 5 shows the various solutions to the differential equations.

Figure 5. Numerous solutions to the differential equation tex:$$\dot{q} + q = 2u(t - 0.5)$$.


You will note that in each case, there are many different solutions to a differential equation, but if we start a circuit with the same conditions at time tex:$$t = 0$$, we expect the response of the circuit to be the same.

In this case, the state of the circuit defines which solution we choose, in particular, the we look at the initial charge on the capacitor.

Initial-value Problems

The previous lecture has shown that given any 1st-order ODE, there exist multiple solutions. When a differential equation is used to describe, for example, a discharging capacitor, the charge on the capacitor follows only one of these solutions. To select one of the many solutions, we must introduce a constraint which is usually of the form

tex:$$y(t_0) = y_0$$,

that is, we know the state of the solution at one particular moment in time and from this, we should be able to deduce the solution for all time. To simplify the mathematics, we will usually define the point in time when the solution is known to be tex:$${t_0 = 0$$, that is, we specify

tex:$$y(0) = y_0$$.

Because we are specifying the state of the system at one point in time and usually we are attempting to determine the solution from that point on, that is, for tex:$$t > t_0$$, we call the constraint an initial value and the ordinary differential equation together with the initial value is called an initial-value problem.

Homogeneous 1st-order Linear Ordinary Differential Equations

A homogeneous 1st-order LODE is of the form

tex:$$\dot{y} + a_0(t)y = 0$$

where the function tex:$$a_0(t)$$ is a function of the independent variable. Suppose we need to find a solution to this differential equation subject to the condition tex:$$y(t_0) = y_0$$.