TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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

givenL1L2L3L4

2.1

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.

step 1.1L2L5L6algebra∎

Depends on

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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)