Differential and Integral Equations: Rates, Accumulation, and Their Connection

How two mathematical viewpoints describe change—and how to move between them in a concrete example.

Differential equations describe an unknown function through its derivatives; integral equations describe it through integrals involving that function. Their useful connection is the relationship between rates of change and accumulation. A worked example below shows how one problem can have both descriptions, while keeping clear the difference between integrating a known expression and solving an integral equation. [3], [2]

In this article

1. Two viewpoints on an unknown function
2. Integrating an equation is not always solving an integral equation
3. Worked example: the same problem in both forms
4. Reading the labels on integral equations
5. Choose a useful form—and verify the answer

1. Two viewpoints on an unknown function

Differential viewpoint: specify how an unknown function changes by relating it to one or more derivatives. [3]
Integral viewpoint: specify a relationship in which the unknown function occurs inside an integral. [2]
The practical connection: integration can reconstruct a function from a rate, but additional information may be needed to select a particular solution. [3]

In both cases, the answer sought is a function, not simply a number. For example, OpenStax introduces the differential equation y′ = 3x²: the task is to find a function whose derivative equals 3x². The function y = x³ works. This illustrates why differential equations are useful descriptions of change: they constrain a function through its behavior rather than directly giving its values. MIT describes ordinary differential equations as involving functions of one variable, often interpreted as time, and emphasizes their importance in science and engineering. [3], [1]

2. Integrating an equation is not always solving an integral equation

Consider the derived example y′ = 3x² with the added condition y(0) = 2. Integration gives y(x) = x³ + C; inserting the condition gives C = 2. Thus y(x) = x³ + 2. Differentiation checks the rate law, and substitution at zero checks the initial value. The constant matters because differentiation erases additive constants: knowing the rate alone does not distinguish these vertically shifted functions. OpenStax makes this point with x² and x² + 4, both of which solve y′ = 2x. [3]

We could also write this calculation as y(x) = 2 + ∫₀ˣ 3t² dt. However, the integrand is already known: it is 3t², not the unknown function y. Under MathWorld’s definition, this is not an integral equation for y merely because it contains an integral. A genuine integral equation places the unknown function inside the integral. This distinction prevents a common confusion: integration is an operation; an integral equation is a type of problem. [3], [2]

3. Worked example: the same problem in both forms

For an original mathematical example, suppose y′(x) = y(x) and y(0) = 1. This specifies a rate equal to the current value of the unknown function. Integrating the rate relationship from zero to x gives y(x) − y(0) = ∫₀ˣ y(t) dt, hence y(x) = 1 + ∫₀ˣ y(t) dt. Unlike the previous example, the unknown is now inside the integral. The differential statement describes the rate at each point; the integral statement describes the value as its starting value plus accumulated change. [3], [2]

Check the candidate y(x) = eˣ in both forms. Its derivative is eˣ, so it satisfies y′ = y, and its value at zero is one. In the integral form, 1 + ∫₀ˣ eᵗ dt = 1 + (eˣ − 1) = eˣ. Conversely, for a continuous function satisfying this integral equation, differentiating gives y′ = y, while setting x = 0 gives the initial value. These calculations establish the connection for this example; they are not a claim that every differential and integral equation can be interchanged without checking conditions. [3], [2]

4. Reading the labels on integral equations

Integral equations have two useful, separate classification questions: Do the integration limits vary? And where does the unknown function appear? MathWorld calls equations with fixed integration limits Fredholm equations, and those with one variable limit Volterra equations. It calls an equation first kind when the unknown appears only under the integral sign, and second kind when it appears both inside and outside. These labels describe structure rather than ranking difficulty, and MathWorld notes that not every integral equation fits the listed standard forms. [2]

The table separates the two classification axes. Our example y(x) = 1 + ∫₀ˣ y(t) dt is Volterra, second kind: its upper limit varies with x, and y appears on both sides of the integral sign. Neither label alone supplies both pieces of information. [2]

Classification axis	Label	What to inspect
Integration limits	Fredholm	The integration limits are fixed. [2]
Integration limits	Volterra	One integration limit is variable. [2]
Location of the unknown	First kind	The unknown appears only inside the integral. [2]
Location of the unknown	Second kind	The unknown appears inside and outside the integral. [2]

5. Choose a useful form—and verify the answer

A practical starting point is to ask what information the problem supplies. A prescribed rate suggests a differential formulation; a relationship involving integrals of the unknown suggests an integral formulation. This is guidance, not a universal rule about which is easier. OpenStax lists direct solution, graphs, and computer calculations as approaches to differential equations. MathWorld identifies Neumann series as a technique for second-kind integral equations. The useful choice therefore depends on the equation’s structure and the methods available, not simply on which notation looks shorter. [3], [2]

Whatever method produces a candidate, check the original problem. OpenStax defines a differential-equation solution through substitution of the function and its derivatives. The analogous check in our integral example is to evaluate the integral and compare both sides. Also check any stated initial value separately unless it is already built into the formulation. This separates finding a plausible expression from demonstrating that it answers the question—and makes the interplay between differentiation and integration a usable calculation rather than just a conceptual analogy. [3], [2]

Frequently asked questions

Is an integral equation just an equation containing an integral?

Not in the relevant sense: the integral must involve the function being solved for. An expression such as y(x) = 2 + ∫₀ˣ 3t² dt reconstructs a function from a known integrand. By contrast, y(x) = 1 + ∫₀ˣ y(t) dt contains the unknown within the integral and is an integral equation. [2]

Why does an initial value matter?

Different functions can share the same derivative because additive constants disappear under differentiation. An initial value can distinguish them: in the example y′ = 3x², specifying y(0) = 2 selects x³ + 2 from the family x³ + C. This example does not establish a universal uniqueness rule for all equations. [3]

What does “ordinary” mean in ordinary differential equation?

MIT describes ordinary differential equations as dealing with functions of one variable, which can often be interpreted as time. In the worked examples, that variable is x. The variable’s name does not determine the classification; the important point is that the unknown function depends on one independent variable. [1]

Does “second kind” mean a second derivative is involved?

No. For integral equations, “second kind” means that the unknown function appears both inside and outside the integral. It describes the placement of the unknown, not a derivative. Separately, Fredholm and Volterra distinguish fixed from variable integration limits. [2]

Must a solution have a simple explicit formula?

The available approaches are broader than finding an elementary expression. OpenStax includes graphs and computer calculations alongside direct solutions. MathWorld also identifies a series method for second-kind integral equations. The appropriate approach depends on the specific equation; the simple exponential example illustrates the connection without representing every problem’s difficulty. [3], [2]

Conclusion

The central connection is straightforward: differential equations constrain rates, while integral equations constrain functions through accumulated contributions. In the worked example, y′ = y with y(0) = 1 becomes y(x) = 1 + ∫₀ˣ y(t) dt, and eˣ satisfies both. To use these ideas, identify the unknown, inspect where it appears, retain the initial information, and verify the candidate in the original equation. Those distinctions are more useful than treating every integration problem as an integral equation. [3], [2]

Source notes

The worked conversion is established for the stated example, not as a general equivalence theorem for arbitrary differential and integral equations.

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