Order and linearity of differential equations
Problem 6.416 · easy
Classify \( \displaystyle 3 y + y' - 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- 3 x} \) 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.
- \[ \frac{d}{d x} \left(1 + e^{- 3 x}\right) + 3 e^{- 3 x} = 0 \]Substituting y = 1 + exp(-3*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. A reviewer's objection to the explanation stands; see 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 substitutes only the derivative and the term 3e^{-3x}, omitting the 3y and -3 terms from the equation. The correct substitution shows 3(1+e^{-3x})+(-3e^{-3x})-3=0, which is not reflected in the solution.qwen3.6:27b-mlx: fail (error) — The verification step is incorrect. Substituting y = 1 + e^{-3x} into the original equation 3y + y' - 3 = 0 yields 3(1 + e^{-3x}) + (-3e^{-3x}) - 3 = 0, which simplifies to 0 = 0. The solution provided computes y' + 3e^{-3x} = 0, which corresponds to a different equation (y' + 3y = 3 is not what was checked; rather, it seems to have checked y' + 3(y-1) = 0 or simply made an algebraic error in substitution). Specifically, the equation shown is Derivative(...) + 3*exp(-3*x) = 0, which is y' + 3(y-1) = 0, not 3y + y' - 3 = 0. While the final result 0=0 is true for the correct substitution, the intermediate equation displayed is not the direct substitution into the given ODE, making the verification misleading/wrongly presented.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The verification step is incorrect. Substituting y = 1 + e^{-3x} into the original equation 3y + y' - 3 = 0 yields 3(1 + e^{-3x}) + (-3e^{-3x}) - 3 = 0, which simplifies to 0 = 0. The solution provided computes y' + 3e^{-3x} = 0, which corresponds to a different equation (y' + 3y = 3 is not what was checked; rather, it seems to have checked y' + 3(y-1) = 0 or simply made an algebraic error in substitution). Specifically, the equation shown is Derivative(...) + 3*exp(-3*x) = 0, which is y' + 3(y-1) = 0, not 3y + y' - 3 = 0. While the final result 0=0 is true for the correct substitution, the intermediate equation displayed is not the direct substitution into the given ODE, making the verification misleading/wrongly presented.gpt-oss:20b: fail (error) 2026-10-17 — The verification step substitutes only the derivative and the term 3e^{-3x}, omitting the 3y and -3 terms from the equation. The correct substitution shows 3(1+e^{-3x})+(-3e^{-3x})-3=0, which is not reflected in the solution.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The verification step is incorrect. Substituting y = 1 + e^{-3x} into 3y + y' - 3 yields 3(1 + e^{-3x}) - 3e^{-3x} - 3 = 0, which is correct. However, the equation shown in line 3 is Derivative(1 + exp(-3*x), x) + 3*exp(-3*x) = 0, which simplifies to -3e^{-3x} + 3e^{-3x} = 0. This equation corresponds to y' + 3(y-1) = 0 or similar, but it omits the constant terms from the original ODE (the '3y' and '-3'). Specifically, the term '3*1' from '3y' and the '-3' from the ODE cancel out, but the line presented does not show this cancellation or the full substitution, making the verification logically incomplete and potentially misleading about what was checked.gpt-oss:20b: fail (error) 2026-10-14 — The verification step substitutes y=1+e^{-3x} into the wrong expression (y'+3e^{-3x}=0) instead of the original differential equation 3y+y'-3=0, so the conclusion that the function satisfies the equation is incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The verification step is algebraically incorrect. Substituting y = 1 + e^{-3x} into 3y + y' - 3 yields 3(1 + e^{-3x}) + (-3e^{-3x}) - 3 = 0, but the solution line shows 3*exp(-3*x) instead of 3*(1 + exp(-3*x)), omitting the constant term from y. This makes the verification invalid.gpt-oss:20b: fail (error) 2026-10-14 — The verification step is incorrect: it omits the constant term and the 3y term, so the substituted expression is not shown to equal zero. The correct check should compute y' + 3y - 3 and show it simplifies to 0.
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 2026-10-17 with SymPy 1.14.0.