A differential equation is an equation involving derivatives. For example, \[ \frac{dy}{dx} - 2y = 0 \quad \] is a differential equation.
Notation
When talking about differential equations it is common to set \[ \frac{dy}{dx} = y \ ' \] so the equation above is written \[ y' - 2y = 0 \]
The equation contains a first derivative only so it is called a first order equation. Another differential equation is \[ \frac{d^2y}{dx^2} + ky = 0 \quad \text{or} \quad y'' + ky = 0 \] The highest derivative in this equation is the second derivative so it is called a second order equation.
Solutions of algebraic equations are sets of numbers; solutions of differential equations are sets of functions.
Verifying Solutions
You verify a given function is a solution of a differential equation by substituting in the equation.
Example To verify that \[ y = \frac{x^2}{3} + \frac{1}{x} \quad \] is a solution to \[ xy' + y = x^2 \qquad (1) \] differentiate y \[ y' = \frac{2x}{3} - \frac{1}{x^2} \] and substitute into (1): \[ \begin{align} xy' + y &= x \left(\frac{2x}{3} - \frac{1}{x^2} \right) + \left(\frac{x^2}{3} + \frac{1}{x} \right) \quad \\ &= x^2 \quad \\ \end{align} \ \] for all x not equal to zero.
Solutions are defined on intervals. In the equation above the solution is defined on any interval \( (a \lt x \lt b ) \) that does not include 0, because y is not defined at 0.
The simplest differential equation has the form \[ \frac{dy}{dx} = f(x) \] You solve this by integrating both sides of the equation with respect to x: \[ y = \int f(x)dx \quad \] and the solution is \[ y = F(x) + c \quad \text{where} \quad F'(x) = f(x) \] from the Fundamental Theorem of Calculus.
A solution containing the arbitrary constant c is called a general solution.
Find the general solution to \(y' = \dfrac{1}{x}\).
The general solution of a differential equation always includes as many constants as the order of the equation. In most cases, the problem is to find a solution that satisfies a specific condition, i.e. passes through a particular point.
Example For the equation \[ y' = 3x^2 \] integrating gives \[ y = x^3 + c \] which is called a general solution because the value of c has not been determined. Now say the solution must satisfy the condition y(1)= 3. Setting x = 1 we get \[ \begin{align} 3 &= 1^3 + c \\ c &= 2 \end{align}\] so a particular solution is \( y = x^3 + 2 \) which passes through the point (1, 3).
The combination of a differential equation with an initial value is called an initial value problem.
Find the solution of y' = 1/x which passes through the point (1,1).
The graph of a particular solution is a solution curve, also called an integral curve.
If \[ y' = f(x, y) \] and f(x, y) is defined on a rectangle R[\( (a \lt x \lt b ) (c \lt y \lt d ) \)], a direction field is a plot of f(x, y) for selected points in R. Here is an example of the direction field for the equation \[ y' = y - x \] The lines show the direction of y' at the point which is at the mid point of the lines. That is, they show the direction of a tangent to a solution curve.
Direction fields are used to investigate the behavior of y without having to solve for y.
Example In the example above, you can see three different behaviors as x increases. If y' is initially one then y increases directly with x along the line y = x + 1. If y' is initially less than one, y becomes large and negative as x increases. If y' is initially greater than one, y becomes large and positive as x increases.
Guided Examples
O E Qs
The graph of a particular solution is a solution curve or an integral curve.
You can sketch solution curves on a direction field. To do this:
Example Here are some solution curves sketched on the direction field above.
In some cases you may be asked for a solution curve that passess through a particular point. This will usually be specified in the form y(x0)= y0, which means the solution passing through (x0, y0). In the diagram above, the top solution curve has initial condition y(0) = 3/2, and passes through the point \( (0, 3/2) \).
Write down an estimate of the initial condition y(0) for the bottom solution curve in the diagram above.
Here is a process for constructing direction fields:
A partial table for the upper right quadrant is shown below.
y\x | 0 | 1 | 2 | 3 | 4 |
0 | 0 | -1 | -2 | -3 | -4 |
1 | 1 | 0 | -1 | -2 | -3 |
2 | 2 | 1 | 0 | -1 | -2 |
3 | 3 | 2 | 1 | 0 | -1 |
4 | 4 | 3 | 2 | 1 | 0 |
To determine the features of a direction field of an equation, substitute x and y values into the equation. Here are some things to look for.
Features are useful for matching direction fields and equations.
In some cases you are given a direction field and some candidate equations for the field.
Here is a process for selecting a candidate equation when you are provided with a direction field:
Example A direction field is shown in the diagram below.
The candidate equations are: \[ \begin{align} \frac{dy}{dx} &= x - y \quad &(A)\\ \frac{dy}{dx} &= \frac{x}{y} \quad &(B)\\ \frac{dy}{dx} &= -xy \quad &(C)\\ \frac{dy}{dx} &= \frac{x}{1+y^2} \quad &(D)\\ \end{align}\]
In the given direction field, y' = 0 when x = 0 and y' = 0 when y = 0. This is probably a defining feature.
For equation A, y' = 0 when y = x. This eliminates A.
For equation B, y' = 1 when y = x (except y = 0). This eliminates B.
For equation C, y' = 0 when x = 0 and y' = 0 when y = 0. So C is a candidate.
For equation D, y' increases directly with x along the x axis.. This eliminates D.
After eliminating equations A, B and D, the only remaining candidate is C.
Write down as many features of the direction field in the example above as you can.