Antiderivatives with initial conditions
Problem 3.479 · hard
Solve \( \displaystyle f''(x) = 2 e^{x} - 2 \) with \( \displaystyle f'(0) = 3 \) and \( \displaystyle f(0) = 1 \).
- \[ \frac{d}{d x} \left(- 2 x + 2 e^{x}\right) = 2 e^{x} - 2 \]An antiderivative of f″; f′ is this plus a constant C₁.✓ Proved
- \[ 3 - \left. - 2 x + 2 e^{x} \right|_{\substack{ x=0 }} = 1 \]f′(0) = 3 fixes C₁.✓ Proved
- \[ \frac{d}{d x} \left(- x^{2} + x + 2 e^{x}\right) = - 2 x + 2 e^{x} + 1 \]Antidifferentiate f′; add a constant C₀.✓ Proved
- \[ 1 - \left. - x^{2} + x + 2 e^{x} \right|_{\substack{ x=0 }} = -1 \]f(0) = 1 fixes C₀.✓ Proved
- \[ - x^{2} + x + 2 e^{x} - 1 \]The solution.✓ Proved
Answer \( f(x) = - x^{2} + x + 2 e^{x} - 1 \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the answer differentiated back to the given derivative, and the initial values checked |
Reviewers
gpt-oss:20b: fail (error) — The solution omits the constant of integration when integrating f'' to obtain f′. Step 1 should produce f′(x)=−2x+2e^x+C₁, and Step 2 must solve for C₁ using f′(0)=3. Because C₁ is missing, the subsequent steps are based on an incorrect f′, leading to an incorrect final function.qwen3.6:27b-mlx: pass
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-07 — The solution omits the constant of integration when integrating f'' to obtain f′. Step 1 should produce f′(x)=−2x+2e^x+C₁, and Step 2 must solve for C₁ using f′(0)=3. Because C₁ is missing, the subsequent steps are based on an incorrect f′, leading to an incorrect final function.qwen3.6:27b-mlx: pass 2026-10-07gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly integrates twice and applies the initial conditions to determine the constants. The algebraic steps and logic are sound.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/antiderivative_ivp, checked 2026-10-07 with SymPy 1.14.0.