Order and linearity of differential equations
Problem 6.570 · easy
Classify \( \displaystyle y + y' - 1 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- x} \) is a solution.
- The highest derivative is y′, so the order is 1.
- It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is nonzero: nonhomogeneous.
- \[ \frac{d}{d x} \left(1 + e^{- x}\right) + e^{- x} = 0 \]Substituting y = 1 + exp(-x) makes the left side 0.✓ Proved
Answer \( \text{order 1, linear and nonhomogeneous} \)
Lines: 1 proved, 2 not checked. 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ 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 | order from sympy.ode_order; linearity from the equation's second partials in y, y′, y″; the solution checked in floating point |
Reviewers
gpt-oss:20b: fail (error) — The verification step omits the constant term –1 from the differential equation, so the substituted expression is not the full left‑hand side. The student would learn an incorrect check of the solution.qwen3.6:27b-mlx: fail (error) — The verification step is incorrect. Substituting y = 1 + e^{-x} into y + y' - 1 yields (1 + e^{-x}) + (-e^{-x}) - 1 = 0, but the solution claims the derivative plus the exponential term equals 0, omitting the constant terms and misrepresenting the substitution process.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The verification step is incorrect. Substituting y = 1 + e^{-x} into y + y' - 1 yields (1 + e^{-x}) + (-e^{-x}) - 1 = 0, but the solution claims the derivative plus the exponential term equals 0, omitting the constant terms and misrepresenting the substitution process.gpt-oss:20b: fail (error) 2026-10-18 — The verification step omits the constant term –1 from the differential equation, so the substituted expression is not the full left‑hand side. The student would learn an incorrect check of the solution.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The verification step is incorrect. Substituting y = 1 + e^{-x} into the ODE y + y' - 1 = 0 yields (1 + e^{-x}) + (-e^{-x}) - 1 = 0, which simplifies to 0 = 0. The provided equation 'Derivative(1 + exp(-x), x) + exp(-x) = 0' simplifies to -e^{-x} + e^{-x} = 0, which is true but does not represent the substitution into the original ODE (it misses the constant terms 1 and -1). This teaches a student that verification involves checking a derived homogeneous equation rather than the original non-homogeneous one.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The verification step is algebraically incorrect. Substituting y = 1 + e^{-x} into y + y' - 1 yields (1 + e^{-x}) + (-e^{-x}) - 1 = 0, but the solution claims the left side is Derivative(1 + exp(-x), x) + exp(-x), which equals 0, ignoring the constant terms 1 and -1 from the original equation y + y' - 1 = 0. The equation shown in step 3 corresponds to y' + y = 0, not the given ODE.gpt-oss:20b: fail (error) 2026-10-15 — The verification step omits the constant term –1 from the differential equation, so the conclusion that the left‑hand side equals zero is not justified.
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/de_classify, checked 2026-10-18 with SymPy 1.14.0.