Order and linearity of differential equations
Problem 6.673 · easy
Classify \( \displaystyle 2 y + y' - 2 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- 2 x} \) 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.
- \[ \frac{d}{d x} \left(1 + e^{- 2 x}\right) + 2 e^{- 2 x} = 0 \]Substituting y = 1 + exp(-2*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 terms 2y and –2 from the differential equation. It only checks that y′+2e^{-2x}=0, which is not equivalent to the original equation 2y+y′-2=0. The conclusion that y=1+e^{-2x} satisfies the equation is therefore unsupported.qwen3.6:27b-mlx: fail (error) — The verification step is incorrect. Substituting y = 1 + e^{-2x} into 2y + y' - 2 yields 2(1 + e^{-2x}) + (-2e^{-2x}) - 2 = 0, which is correct. However, the equation shown in line 3 is 'Derivative(...) + 2*exp(-2*x) = 0', which simplifies to -2e^{-2x} + 2e^{-2x} = 0. This equation misses the constant terms (2y - 2) from the original ODE, effectively verifying y' + 2e^{-2x} = 0 instead of the full ODE. While the result 0=0 is true, the expression shown does not represent the substitution into the given ODE 2y + y' - 2 = 0.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The verification step is incorrect. Substituting y = 1 + e^{-2x} into 2y + y' - 2 yields 2(1 + e^{-2x}) + (-2e^{-2x}) - 2 = 0, which is correct. However, the equation shown in line 3 is 'Derivative(...) + 2*exp(-2*x) = 0', which simplifies to -2e^{-2x} + 2e^{-2x} = 0. This equation misses the constant terms (2y - 2) from the original ODE, effectively verifying y' + 2e^{-2x} = 0 instead of the full ODE. While the result 0=0 is true, the expression shown does not represent the substitution into the given ODE 2y + y' - 2 = 0.gpt-oss:20b: fail (error) 2026-10-19 — The verification step omits the terms 2y and –2 from the differential equation. It only checks that y′+2e^{-2x}=0, which is not equivalent to the original equation 2y+y′-2=0. The conclusion that y=1+e^{-2x} satisfies the equation is therefore unsupported.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The verification step is incorrect. Substituting y = 1 + e^{-2x} into 2y + y' - 2 yields 2(1 + e^{-2x}) + (-2e^{-2x}) - 2 = 0, which is correct. However, the equation shown in line 3 is y' + 2e^{-2x} = 0, which omits the constant terms from 2y and -2, failing to actually verify the original ODE.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-16 — The verification step is incorrect. Substituting y = 1 + e^{-2x} into 2y + y' - 2 yields 2(1 + e^{-2x}) + (-2e^{-2x}) - 2 = 0, which is correct. However, the equation shown in line 3 is 'Derivative(...) + 2*exp(-2*x) = 0', which corresponds to y' + 2e^{-2x} = 0. This ignores the '2y' and '-2' terms from the original ODE, effectively verifying a different equation (y' + 2e^{-2x} = 0) rather than the given one. The algebraic check presented does not match the problem statement.gpt-oss:20b: fail (error) 2026-10-16 — The verification step omits the 2y term; it only checks y' + 2e^{-2x}=0, which does not correspond to the original equation 2y + y' - 2 = 0. The student would learn an incorrect substitution test.
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 2026-10-19 with SymPy 1.14.0.