Order and linearity of differential equations
Problem 6.729 · 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. 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 is incorrect: it only substitutes y' and the exponential term, omitting the 3y and constant 3 terms from the differential equation. The calculation does not actually show that y = -1 + e^{-3x} satisfies 3y + y' + 3 = 0.qwen3.6:27b-mlx: pass — The classification is correct: the equation is first-order, linear (y and y' are to the first power), and nonhomogeneous (due to the constant term 3). The verification step correctly substitutes the proposed solution into the differential equation and confirms it satisfies the equation.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-25 — The classification is correct: the equation is first-order, linear (y and y' are to the first power), and nonhomogeneous (due to the constant term 3). The verification step correctly substitutes the proposed solution into the differential equation and confirms it satisfies the equation.gpt-oss:20b: fail (error) 2026-10-25 — The verification step is incorrect: it only substitutes y' and the exponential term, omitting the 3y and constant 3 terms from the differential equation. The calculation does not actually show that y = -1 + e^{-3x} satisfies 3y + y' + 3 = 0.qwen3.6:27b-mlx: pass 2026-10-24 — The classification is correct and the verification step correctly substitutes the proposed solution into the differential equation.gpt-oss:20b: fail (error) 2026-10-24 — The verification step omitted the 3y term; it only checked y' + 3e^{-3x}=0, which is not the original differential equation. The correct substitution should evaluate 3(-1+e^{-3x}) + (-3e^{-3x}) + 3 and show it equals 0.qwen3.6:27b-mlx: fail (error) 2026-10-24 — The verification step is incorrect. The substitution of y = -1 + e^{-3x} into the equation 3y + y' + 3 = 0 yields 3(-1 + e^{-3x}) + (-3e^{-3x}) + 3 = -3 + 3e^{-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 checking if y' + 3(y+1) = 0 or similar, but it misses the constant terms from 3y and +3. Specifically, 3y = 3(-1 + e^{-3x}) = -3 + 3e^{-3x}. The line shown effectively checks -3e^{-3x} + 3e^{-3x} = 0, ignoring the -3 from 3y and the +3 constant. While the final result 0=0 is true, the intermediate equation displayed is not the direct substitution of the original ODE. It looks like the solver substituted y' + 3y + 3 but somehow dropped the constants or rearranged incorrectly in the display. Wait, let's look closer. The line says: Derivative(...) + 3*exp(-3*x) = 0. The derivative is -3e^{-3x}. So -3e^{-3x} + 3e^{-3x} = 0. This is 0=0. But the original equation is 3y + y' + 3 = 0. Substituting y gives 3(-1 + e^{-3x}) + (-3e^{-3x}) + 3 = -3 + 3e^{-3x} - 3e^{-3x} + 3 = 0. The line shown does not represent the full substitution. It represents a partial check or a miscalculation of the substitution process. A student would be misled into thinking that substituting y into 3y+y'+3 results in just y' + 3e^{-3x} or similar. The line is factually wrong as a representation of the verification step.gpt-oss:20b: fail (error) 2026-10-24 — The verification step only computes y' + 3e^{-3x} and sets it to zero, omitting the 3y + 3 terms from the differential equation. Thus the solution is not properly verified.
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 2026-10-25 with SymPy 1.14.0.