What Is a Differential Equation?

Most equations you have met so far ask for a number: solve x^2 = 9 and you get x = \pm 3. A differential equation raises the stakes — it asks for a whole function, and it pins that function down not by its values directly but by a relationship between the function and its derivatives.

For instance, "find a function that is equal to its own rate of change" is the equation

\frac{dy}{dx} = y.

The unknown is the function y(x), and the equation is a constraint it must satisfy at every point. The whole subject is the art of turning such constraints back into explicit functions.

Order — how many derivatives deep

The order of a differential equation is the highest derivative that appears in it. It is the single most useful label, because the solution method follows from it.

y' = 3x^2 is first order — only y' appears.
y'' + 4y = 0 is second order — the highest is y'' (the 4y term, being order zero, does not raise it).
(y'')^3 + y' = x is still second order: order counts the highest derivative present, not the power it is raised to.

An equation involving a function of one variable and its ordinary derivatives is an ODE (ordinary differential equation). When the unknown depends on several variables and the equation involves partial derivatives — like the heat equation \partial_t u = \partial_{xx} u — it is a PDE. This page, and everything that follows, is about ODEs.

A solution is a function — verify one, line by line

A solution of an ODE is a function that, when substituted in along with its derivatives, makes the equation an identity — true for all x in some interval. Checking a candidate is pure substitution; no cleverness required.

Step 1 — state the claim. We claim y = e^{2x} solves the first-order ODE

y' = 2y.

Step 2 — differentiate the candidate. By the chain rule,

y = e^{2x} \;\Longrightarrow\; y' = 2e^{2x}.

Step 3 — substitute both sides. The left side of the ODE is y' = 2e^{2x}. The right side is 2y = 2e^{2x}.

\underbrace{2e^{2x}}_{y'} \;=\; \underbrace{2e^{2x}}_{2y}. \quad\checkmark

Step 4 — conclude. The two sides are identically equal for every x, so y = e^{2x} is indeed a solution. Notice we never had to solve the equation — only confirm a guess, which is always within reach.

General vs particular: the constant of integration

A first-order ODE rarely has just one solution. The very simplest, y' = f(x), is solved by integrating, and integration always trails a constant:

y' = 2x \;\Longrightarrow\; y = \int 2x \, dx = x^2 + C.

Every value of C gives a genuine solution, so what we really have is a whole family — the general solution y = x^2 + C, one curve for each height. To single out one member we impose an initial condition, a known value of the function at one point.

Step 1 — start from the general solution. y = x^2 + C.

Step 2 — impose the initial condition y(1) = 5. Substitute x = 1, y = 5:

5 = (1)^2 + C = 1 + C.

Step 3 — solve for the constant. C = 4.

Step 4 — write the particular solution. The single curve through the point (1, 5) is

y = x^2 + 4.

An ODE plus an initial condition is an initial-value problem (IVP). Geometrically, the general solution is a family of curves filling the plane, and the initial condition picks the one passing through a chosen point.

For an ordinary differential equation in an unknown function y(x):

its order is the highest derivative of y that appears;
a solution is a function that, with its derivatives substituted in, satisfies the equation identically on an interval;
the general solution of an n-th order ODE carries n arbitrary constants; fixing them yields a particular solution;
an initial-value problem pairs the ODE with n conditions (e.g. y(x_0), y'(x_0), \dots) that determine those constants — typically picking out exactly one solution.

We have been blithely assuming an IVP has one solution. It usually does, and the Picard–Lindelöf theorem says exactly when. Consider the first-order IVP

y' = f(x, y), \qquad y(x_0) = y_0.

If f is continuous in a rectangle around (x_0, y_0) and is Lipschitz continuous in y there — meaning there is a constant L with |f(x, y_1) - f(x, y_2)| \le L\,|y_1 - y_2| — then there is an interval around x_0 on which a solution exists and is unique.

The Lipschitz condition is doing real work. Drop it and uniqueness can fail: the IVP y' = \sqrt{|y|}, y(0) = 0 is solved both by y \equiv 0 and by y = \tfrac14 x^2 (for x \ge 0) — two solutions from one start, because \sqrt{|y|} has an infinite slope at y = 0 and so is not Lipschitz there. Continuity alone (Peano's theorem) buys existence but not uniqueness.

The slope field: see every solution at once

Here is the picture that makes ODEs intuitive. The equation y' = f(x, y) hands you the slope of the solution at every point (x, y) — before you have solved anything. Draw a tiny dash with that slope at a grid of points and you get a slope field: a sea of little arrows the solution curves must flow along, like iron filings tracing a magnetic field.

Below is the field for y' = x - y (a first-order linear ODE). Slide the starting height y_0: the highlighted curve is the particular solution of the IVP y' = x - y, y(x_0) = y_0, threaded through the slope field. Every dash it passes through points exactly along it.