Difficulty: Introductory | Prerequisites: Basic calculus, continuity of functions, familiarity with standard ODE form.
Tags: existence theorem, uniqueness theorem, first-order linear ODE, standard form, p(t), g(t), continuity, interval of existence, MATH 441, ordinary differential equations
This is one of the first theoretical tools you meet in an ODE course, and it answers a question students often skip past: before you solve an equation, can you guarantee a solution even exists? The existence and uniqueness theorem for first-order linear ODEs gives you a clean, checkable condition. If you can show that the coefficient functions in the standard form are continuous on an interval containing your initial point, the theorem promises exactly one solution on that interval. You will need this both for justifying answers on proofs and for identifying valid solution intervals on applied problems.
Rewrite the ODE in standard form y' + p(t)y = g(t). If p(t) and g(t) are both continuous on an open interval containing the initial point, then a unique solution exists on that entire interval. Your job is to find that interval by identifying where continuity breaks down.
Standard form (first-order linear ODE)
y' + p(t)y = g(t). Every first-order linear ODE can be rearranged into this shape. Think of it as the "normal form" that lets you read off p(t) and g(t) directly.
Existence and uniqueness theorem (for linear ODEs)
If p(t) and g(t) are continuous on an open interval I containing the initial point t₀, then the IVP y' + p(t)y = g(t), y(t₀) = y₀ has exactly one solution, and that solution is valid on the entire interval I. In simple terms, continuity of the coefficients on an interval around your starting point is all you need.
Interval of existence
The largest open interval containing the initial point on which both p(t) and g(t) remain continuous. This is the interval you report. Think of it as the safe zone where nothing blows up.
Start from whatever form the ODE is given in and divide or rearrange until you isolate y' with coefficient 1.
Everything multiplying y becomes p(t). Everything on the right-hand side becomes g(t).
Example: ln(t) y' + y = cot(t) divides through by ln(t) to give y' + (1/ln t) y = cot(t)/ln(t), so p(t) = 1/ln(t) and g(t) = cot(t)/ln(t).
List every value of t that makes a denominator zero, a logarithm undefined, or a trigonometric function undefined.
For p(t) = 1/ln(t): undefined at t ≤ 0 (ln not defined) and at t = 1 (ln 1 = 0).
For g(t) = cot(t)/ln(t): cot(t) = cos(t)/sin(t) is undefined at t = kπ for any integer k, and the ln(t) denominator adds the same restrictions as above.
Collect all these bad points into a single list.
The interval must be open, must contain the initial point t₀, and must not contain any of the discontinuity points.
Walk outward from t₀ in both directions until you hit the nearest discontinuity on each side. Those become the endpoints.
Example: with t₀ = 2 and discontinuities at t = 0, t = 1, and t = π ≈ 3.14, the interval is (1, π).
Once the interval is identified, cite the theorem: "p(t) and g(t) are continuous on (1, π), which contains t₀ = 2, so by the existence and uniqueness theorem for linear ODEs, a unique solution exists on (1, π)."
Standard form:
y' + p(t) y = g(t)
Theorem statement (compact):
If p, g continuous on open interval I containing t₀, then the IVP y' + p(t)y = g(t), y(t₀) = y₀ has a unique solution on all of I.
This theorem underpins every numerical ODE solver's validity check. Before an engineer trusts a simulation of, say, a circuit's transient response (modelled by a first-order linear ODE), they need to know the solution exists and is unique on the time window they care about. Without that guarantee, the solver's output could be meaningless.
Students often choose a closed interval like [1, π]. The theorem requires an open interval, because continuity at the endpoints is not guaranteed.
Some students forget that t must be positive for ln(t) to exist. They write discontinuities at t = 1 and t = kπ but ignore t ≤ 0 entirely.
A common mistake is picking the largest possible interval rather than the one containing the initial point. If t₀ = 2, the interval (π, 2π) is also free of discontinuities, but it does not contain t₀ = 2, so it is irrelevant.
Students sometimes try to solve the ODE first and then find the interval. The point of the theorem is that you can determine existence without solving.
⚠️ Exam problems almost always require you to state the interval explicitly and cite the theorem. Simply saying "a solution exists" without naming the interval will lose marks.
⚠️ Watch for ODEs where the leading coefficient is not 1. You must divide through to reach standard form before identifying p(t) and g(t).
⚠️ If the problem says "determine an interval," it is asking for existence/uniqueness analysis, not for solving the ODE.
True or false: The existence and uniqueness theorem guarantees a solution on a closed interval.
Fill in the blank: To apply the theorem, you must first write the ODE in ________ form.
True or false: If p(t) is continuous on (0, 5) but g(t) has a discontinuity at t = 3, and t₀ = 2, you can still use the interval (0, 5).
Fill in the blank: The interval must be open, must contain ________, and must avoid all points where p or g are discontinuous.
Answers: 1. False (open interval). 2. Standard. 3. False (you must shrink to (0, 3)). 4. The initial point t₀.
Q: Given ln(t) y' + y = cot(t), y(2) = 3, identify p(t) and g(t) in standard form.
A: Divide by ln(t): y' + (1/ln t) y = cot(t)/ln(t). So p(t) = 1/ln(t), g(t) = cot(t)/ln(t).
Q: For the same ODE, list all points where p(t) or g(t) are discontinuous.
A: p(t) is discontinuous at t ≤ 0 and t = 1. g(t) adds discontinuities at t = kπ (integer k). Combined: t ≤ 0, t = 1, t = π, t = 2π, etc.
Q: What is the interval of existence for y(2) = 3?
A: The nearest discontinuities to t = 2 are t = 1 (left) and t = π (right). The interval is (1, π).
Q: Why is the interval open rather than closed?
A: The theorem requires p(t) and g(t) to be continuous at every point in the interval. At the boundary points (t = 1 and t = π), one or both functions are undefined, so those points cannot be included.
This connects directly to the integrating factor method: once the theorem guarantees a solution exists, the integrating factor technique is how you find it. The analogous theorem for nonlinear ODEs (Picard-Lindelöf) has stricter hypotheses, which is why the linear version is singled out as a cleaner result. You will revisit existence and uniqueness again for higher-order linear ODEs later in the course, where the same principle extends to n coefficient functions.
existence and uniqueness, Picard-Lindelöf, first-order linear ODE, standard form ODE, interval of existence, continuity condition, initial value problem, IVP, p(t) and g(t), MATH 441 exam 1, ordinary differential equations, coefficient continuity