TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Existence and uniqueness for y''=-y with prescribed initial data

Statement

For $a, b \in R$ , the function $y (x) = a \cos x + b \sin x$ is the unique twice differentiable function on $R$ satisfying $y^{''} = - y, y (0) = a, y^{'} (0) = b .$

Facts & Assumptions

Given: Reals $a, b$ and a twice differentiable solution $z$ of the displayed initial-value problem.

[L1]

$(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , $\sin 0 = 0$ , and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine).

[L2]

The algebra of derivatives differentiates sums and products (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 $g (c) \neq 0$ ).

[L3]

A differentiable function with zero derivative on $R$ is constant (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

Proof

technique · direct

1.1L1algebra

By [L1], $y = a \cos + b \sin$ satisfies $y^{''} = - y$ , $y (0) = a$ , and $y^{'} (0) = b$ .

2.1step 1.1L2algebra

Put $h = z - y$ . Then $h^{''} = - h$ , $h (0) = h^{'} (0) = 0$ , and $E : = h^{2} + (h^{'})^{2}$ has derivative $E^{'} = 2 h (h + h^{''}) = 0$ .

3.1step 2.1L3algebra∎

Thus $E$ is constantly $E (0) = 0$ ; as both squares are nonnegative, $h = 0$ everywhere and $z = y$ .

Existence and uniqueness for y''=-y with prescribed initial data

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources