Ordinary differential equations

Read a constraint on an unknown function

An algebraic calculation such as 3a + 2 = 8 asks for a number a. A differential equation involves an unknown function and one or more of its derivatives. Its solution is a function that satisfies the equation on a stated interval when the function and its derivatives are substituted.

An ordinary differential equation, abbreviated ODE, uses derivatives with respect to one independent variable. For y(t), t is that input and y the dependent quantity. The equation y′(t) = F(t, y(t)) prescribes a rate through a supplied formula F; it does not yet give y(t) explicitly. The prime means derivative with respect to t.

Although F has places for both t and the current y value, the unknown y(t) still has one independent input. Compare directly supplying x(t) with supplying x′(t): the latter constrains how x changes and may leave an initial value to be chosen. A differential equation alone need not select one function.

Declare quantities, units and the allowed interval

For an original exact teaching model, track one chosen point along a straight line. Fix the origin and take right positive. Let t be time in seconds from a declared clock origin and x(t) signed position in metres. Supply x′(t) = 1.20 m/s + (0.40 m/s²)(t − 1 s), only for 0 < t < 4 s.

Here the unknown is x; the independent variable is t. Its derivative has units m/s. In the right-hand side, (m/s²)s = m/s, so the terms can be added. This is a supplied rate rule at every time in the interval, not a derivative inferred from a few samples. Its numerical coefficients do not claim measurement precision or a force law.

Also supply x(1 s) = −0.50 m. The datum lies inside the open time interval. Write h = t − 1 s, a time offset with second units. Calling the datum initial does not require that its time be zero or the first allowed time: it fixes the solution at the selected reference time.

Identify the order and the initial data

The order is the highest derivative order appearing in the differential equation. Our x′ rule is first order. A separate equation x″ = 0.40 m/s² is second order because x″ is the derivative of x′ with respect to time. The power of t in an explicit polynomial solution does not determine differential order.

For that separate second-order model on 0 < t < 4 s, integration gives x = A + B h + (0.20 m/s²)h². The constants A and B have units m and m/s. Position x(1 s) specifies A; velocity x′(1 s) specifies B. Two different B values can have the same initial position but different later positions.

An equation together with specified initial data is an initial-value problem. In these regular first- and second-order examples, the appropriate values select the constants. Counting data alone is not a universal proof of existence or uniqueness; the equation, its domain and further conditions must also be checked.

Check both the equation and the datum

To test a proposed first-order solution, confirm its domain and differentiability, calculate its derivative, and substitute into the equation at every time in the claimed interval. Then check the initial value separately. An agreement at only one time cannot establish an identity throughout an interval.

For the position problem, consider g(t) = −0.50 m + (1.20 m/s)h + (0.20 m/s²)h². Since h = t − 1 s, its derivative is g′ = 1.20 m/s + (0.40 m/s²)h. It matches the supplied rate throughout 0 < t < 4 s, and h = 0 gives g(1 s) = −0.50 m.

Adding 1 m to g leaves its derivative unchanged, so it still solves the differential equation; it fails the original initial datum. Conversely, omitting the h² term preserves the datum and even the correct rate at 1 s, but gives a constant 1.20 m/s derivative. At 2 s the prescribed rate is 1.60 m/s, so this linear candidate fails the equation.

Integrate a rate that is already known as a function of time

When y′ = f(t) with continuous f on an interval containing t₀, the initial-value solution is y(t) = y₀ + ∫_t₀^t f(s) ds. The letter s is a temporary integration variable. The fundamental theorem gives the derivative f at interior times, and the equal bounds at t = t₀ give the datum.

For our rate, integration gives x = C + (1.20 m/s)h + (0.20 m/s²)h². The factor 0.20 is half the rate coefficient 0.40, because differentiating h² gives 2h. At h = 0 the initial datum sets C = −0.50 m. Every term has metre units.

At t = 3 s, h = 2 s, so x = −0.50 m + 2.40 m + 0.80 m = +2.70 m. This is signed position relative to the chosen origin; the displacement since 1 s is +3.20 m. The continuous rate and datum determine this exact result within the supplied model interval.

Solve a rate that depends on the current state

Now supply a separate hypothetical temperature model on 0 < t < 7 s. Let T_ref = 280 K be a fixed reference temperature and θ(t) = T(t) − T_ref a temperature excess in kelvin. Supply θ′ = −θ/τ, with τ = 2 s, and θ(1 s) = θ₀ = 8 K. The right-hand side has units K/s. This is a stipulated equation, not evidence that every cooling object follows it.

The rate depends on the unknown θ, so it is not a supplied time-only integrand. This equation is separable: on a branch where the dependent value is nonzero, division can separate the state and time factors. First check constant solutions, because division can remove them. Here θ = 0 is a solution of the equation, but it fails the specified 8 K datum.

For the positive branch, put z = θ/θ₀, a positive dimensionless ratio, so z′ = −z/τ and z(1 s) = 1. The natural logarithm ln is the inverse of the positive real exponential; its input here is z, not a kelvin-valued quantity. Its derivative along the solution is (ln z)′ = z′/z = −1/τ. Integrating from 1 s gives ln z = −h/τ, hence θ = (8 K)e^(−h/τ).

The exponent −h/τ is dimensionless. Differentiating this candidate gives θ′ = −(8 K/τ)e^(−h/τ) = −θ/τ; at h = 0 it gives 8 K. It remains positive at every allowed time, justifying the chosen logarithm branch. The explicit check confirms the solution without claiming that all ODEs admit this separation method.

State a sufficient local existence-and-uniqueness test

For y′ = F(t, y), y(t₀) = y₀, use this sufficient local test: F and its state partial derivative ∂F/∂y exist and are continuous on an open rectangle containing (t₀, y₀). The state partial holds t fixed. Then a unique differentiable solution exists on a sufficiently small interval containing t₀, with its graph inside that rectangle. Existence means there is a solution; uniqueness concerns the same equation and the same initial datum.

A bounded continuous state derivative controls nearby rate differences under these regularity conditions, explaining the role of the state derivative in a local uniqueness result. Our position rate has state partial 0 s⁻¹, and the temperature-excess rate has state partial −1/τ = −0.50 s⁻¹. Both rate formulas and these partials are continuous near their data.

A local guarantee does not specify the whole future interval, give a closed formula or validate a physical model. If this sufficient test cannot be applied, that alone gives no automatic verdict of nonexistence or nonuniqueness. Those questions require another argument; a failed sufficient condition is not the opposite conclusion.

Check continuation and the model's limits

Consider a separate formal example with dimensionless unknown z: z′ = z²/τ, τ = 2 s, z(1 s) = 1. Its rate has units s⁻¹ and its state partial 2z/τ is continuous, so the local test applies near the datum. Nevertheless, its initial-value solution is z(t) = 1/[1 − (t − 1 s)/τ] on the connected interval containing the datum and lying below t = 3 s.

At 2 s the solution is 2; at 2.5 s it is 4. As t approaches 3 s from below, the positive denominator approaches zero and z grows without bound. No finite value at 3 s can make a continuous differentiable continuation through that time. Algebraically evaluating the same expression at 4 s gives −2, but that belongs to a separate interval across the singular time, not a continuation of the initial-value solution.

This formal divergence is a mathematical model result, not evidence of a physical quantity becoming infinite. For the position and temperature examples, retain their declared intervals, quantity meanings and supplied assumptions even when their formulas can be evaluated elsewhere. The next lesson, Partial differential equations and Green functions, introduces equations where the unknown depends on multiple independent variables and initial or boundary data require further care.

Worked examples

Try it yourself

1. The supplied equation x′ = 1.20 m/s + (0.40 m/s²)(t − 1 s) has a quadratic solution. What is its differential order? What is the order and derivative unit of the separate equation x″ = 0.40 m/s²?

   Show the answer and reasoning

   The first equation is first order. The separate x″ equation is second order, and x″ has units m/s².

   Order follows the highest derivative in the equation, not the power of t in its solution. The time derivative of position has m/s units; differentiating it again divides by another second.

2. For the original position problem, test q(t) = −0.50 m + (1.20 m/s)(t − 1 s). It matches x(1 s). Does it solve the differential equation on 0 < t < 4 s?

   Show the answer and reasoning

   No. Its derivative is 1.20 m/s, but the prescribed rate at 2 s is 1.60 m/s.

   It passes the datum and the rate at 1 s only. At 2 s the derivative differs by −0.40 m/s from the required rate. A solution must satisfy the rate equation throughout the claimed interval.

3. Keep the original position-rate equation but replace its datum by x(1 s) = +0.50 m. What changes in the solution, and what is x(3 s)?

   Show the answer and reasoning

   The position increases by a constant 1 m at every allowed time, with the same derivative. The new x(3 s) is +3.70 m.

   Only C changes from −0.50 m to +0.50 m. At 3 s the rate-generated terms remain 2.40 m + 0.80 m. The different initial-value problems can have different solutions without violating uniqueness for any one datum.

4. For the stipulated temperature model, find θ and T at t = 1 s + (2 s)ln 4. Can the real exponential's argument be written as −h with h carrying seconds?

   Show the answer and reasoning

   θ = 2 K and T = 282 K. The argument is −h/(2 s), a dimensionless ratio.

   At that time e^(−h/(2 s)) = e^(−ln 4) = 1/4, giving (8 K)/4 = 2 K. The target lies inside 0 < t < 7 s. A unit-bearing time alone is not the stated real exponential input; the time scale removes the unit.

5. Change the temperature-excess datum to θ(1 s) = 0 K while keeping θ′ = −θ/(2 s). Is θ(t) = 0 K a solution? Would dividing by θ or normalizing by this zero datum be valid?

   Show the answer and reasoning

   The constant zero-excess function is a solution. Division by θ or the zero initial datum is invalid.

   Its derivative is 0 K/s and the right-hand side is also 0 K/s. It gives T = T_ref = 280 K, not absolute zero. The positive-branch logarithm method excluded zero during division, so that case must be checked directly. The local uniqueness conditions still hold for this linear rate near the zero datum.

6. For the formal z′ = z²/(2 s), z(1 s) = 1 problem, does the local sufficient test guarantee a finite solution through 3 s? Can the algebraic value −2 at 4 s be joined to the original branch across 3 s?

   Show the answer and reasoning

   No to both. The local test guarantees a solution near 1 s; the checked branch becomes unbounded as t approaches 3 s from below.

   The rate and its state partial are continuous near the initial datum, but the local theorem promises no unlimited interval. At 3 s the solution formula's denominator vanishes, and no finite continuous value can bridge the divergence. The 4 s algebraic value lies on a disconnected branch and does not continue the given initial-value solution.

What to carry forward

• An ODE constrains an unknown function and its ordinary derivatives with respect to one independent variable; it does not merely solve for one number.
• State the dependent quantity, derivative units, initial data and allowed interval before interpreting a solution.
• Check a proposed function against the equation throughout its claimed interval and against the initial datum separately.
• A continuous known time rate can be integrated and its constant selected by initial data. For separable state-dependent equations, check constant solutions before dividing.
• Real logarithm inputs and exponential arguments in these unit-bearing models are dimensionless; halving a temperature excess does not halve absolute temperature.
• Continuous rate and state partial provide a sufficient local uniqueness guarantee. Continuation, all-time existence and physical validity remain separate questions.