TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21

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A scalar first-order linear ODE has a unique solution given by the integrating-factor formula

Statement

Let $I \subseteq R$ be order-convex with at least two elements, let $x_{0} \in I$ , and let $p, q : I \to R$ be continuous. The IVP $y^{'} + p y = q$ , $y (x_{0}) = y_{0}$ , has exactly one solution on $I$ , namely

$y (x) = \exp (- P (x)) (y_{0} + \int_{x_{0}}^{x} \exp (P (t)) q (t) d t), P (x) = \int_{x_{0}}^{x} p (t) d t .$

Facts & Assumptions

Given: The continuous coefficients and initial data in the Statement.

[L1]

The exponential satisfies $(\exp)^{'} = \exp$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[L2]

For a continuous $f$ on an order-convex interval with at least two elements, $x \mapsto \int_{x_{0}}^{x} f$ is a primitive of $f$ . If $a < b$ in the interval and $G$ is any primitive, then $\int_{a}^{b} f = G (b) - G (a)$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[L6]

Oriented integrals satisfy $\int_{b}^{a} f = - \int_{a}^{b} f$ and $\int_{a}^{a} f = 0$ (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L3]

If $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L4]

If $f$ and $g$ are differentiable at $c$ , then $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L5]

For every real $u$ , $\exp (u) > 0$ and $\exp (- u) = 1 / \exp (u)$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Proof

technique · direct

1.1givenL1L2L3L4

By the existence clause of [L2], $P^{'} = p$ ; hence [L3], [L4], and [L1] give $(\exp (P) y)^{'} = \exp (P) (y^{'} + p y) = \exp (P) q$ .

2.1step 1.1L2L5L6algebra∎

If $x_{0} < x$ , apply the evaluation clause of [L2] on $[x_{0}, x]$ ; if $x < x_{0}$ , apply it on $[x, x_{0}]$ and reverse the integral with [L6]; equality is immediate at $x = x_{0}$ . In every case, using $P (x_{0}) = 0$ and dividing by [L5] yields the displayed formula. Direct differentiation verifies it and its initial value, while applying step 1.1 to the difference of two solutions makes that difference zero.

Depends on

Used by

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Ch. 2 (standard reference, not scraped)
Jiri Lebl, Basic Analysis I, Section 6.3 (standard reference, not scraped)