Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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The exponential is the unique solution of y=yy'=y with y(0)=1y(0)=1

Statement

If y:RRy:\mathbb R\to\mathbb R is differentiable, y=yy'=y, and y(0)=1y(0)=1, then y=expy=\exp.

Facts & Assumptions

Given: A differentiable solution yy of the initial-value problem.

[L1]

exp=exp\exp'=\exp, exp(x)=1/exp(x)\exp(-x)=1/\exp(x), and the series definition gives exp(0)=1\exp(0)=1 (The exponential function is smooth and (exp)=exp(\exp)'=\exp, The exponential is positive and satisfies exp(x)=1/exp(x)\exp(-x)=1/\exp(x), The real exponential function and the number ee by a power series).

Proof

technique · direct
1.1

Define h(x)=y(x)exp(x)h(x)=y(x)\exp(-x). By [L1] and [L2], h(x)=y(x)exp(x)y(x)exp(x)=0h'(x)=y'(x)\exp(-x)-y(x)\exp(-x)=0.

givenL1L2
2.1

The differentiable function hh is continuous, so [L3] makes it constant; h(0)=y(0)exp(0)=1h(0)=y(0)\exp(0)=1.

step 1.1L1L3
3.1

Thus y(x)exp(x)=1y(x)\exp(-x)=1, and multiplying by exp(x)\exp(x) gives y(x)=exp(x)y(x)=\exp(x).

step 2.1L1algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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