Differential equations are key to solving electromagnetic and electronic problems.
While most problems are too complex to be worth solving analytically, the concepts are worth knowing.
A differential equation is an equation which relates a function to its derivatives, where the solution is the function itself.
Contents
Classification
There are a few broad categories of differential equations and there is a set of terminology to classify them.
Ordinary
An ordinary differential equation (ODE) describes a function that depends on a single variable.
For example, y = f(x) depends only on x.
Some example ODE’s:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 0$$
$$(1 – x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 3y$$
Partial
An partial differential equation (PDE) describes a function that depends on more than one variable.
For example, y = f(x, t) depends on both x and t. Because it is multivariable a PDE involves partial derivatives.
Some example PDE’s:
$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$
$$\frac{\partial^2 u}{\partial x^2} + 3\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} = \sin{x}$$
$$\frac{\partial^2 u}{\partial t^2} = c^2(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2})$$
Order
The order of an ODE refers to the highest power derivative.
For a PDE the definition is the same but it is the highest power of a partial derivative.
First-order ODE:
$$\frac{\mathrm{d}T}{\mathrm{d}t} = k(A – T)$$
Third-order ODE:
$$\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 3\frac{\mathrm{d}^2y}{\mathrm{d}x^2} – 10\frac{\mathrm{d}y}{\mathrm{d}x} = 0$$
Second-order PDE:
$$\frac{\partial^2 V}{\partial x^2} = LC\frac{\partial^2 V}{\partial t^2}$$
Solutions
Solving differential equations occurs in several steps.
General Solution
As mentioned above the solution to a differential equation is a function.
However because of integrating constants there is actually a family of solutions to each differential equation.
This family is given by what’s known as a general solution.
For instance the discharge of capacitor voltage in an RC-circuit is given by a first-order ODE.
We can write KVL, noting that the current is related to the voltage by the standard capacitor equation:
$$V_C + Ri_C = 0$$
$$V_C = -Ri_C = -RC\frac{\mathrm{d}V_C}{\mathrm{d}t}$$
This equation can be solved by integration, but first all terms involving the unknown function (capacitor voltage) need to be grouped together.
This can be accomplished by moving everything to one side of the equation.
$$\frac{1}{V_C}\frac{\mathrm{d}V_C}{\mathrm{d}t} = -\frac{1}{RC}$$
$$\int\frac{1}{V_C}\frac{\mathrm{d}V_C}{\mathrm{d}t} = -\int\frac{1}{RC}$$
$$\int\frac{1}{V_C}\frac{\mathrm{d}V_C}{\mathrm{d}t} = -\int\frac{1}{RC}$$
$$\ln{V_C} = -\frac{t}{RC} + c_1$$
$$V_C = e^{-\frac{t}{RC} + c_1}$$
$$V_C = e^{c_1}e^{-\frac{t}{RC}}$$
$$V_C = Ae^{-\frac{t}{RC}}$$
Therefore, the solution to the ODE is the function VC(t) but we are stuck with an unknown constant A.
Initial Values
The coefficient A can only be solved if further information is provided which is known as an initial value.
In this case it turns out that A is the capacitor voltage at time 0:
$$V_C(0) = Ae^{0} = A$$
So, if the capacitor started at 9 V at time 0 then A = 9 V.
In general the number of initial values required is equal to the order of the equation.