Order and linearity of differential equations
Problem 6.561 · easy
Classify \( \displaystyle 3 y + y' - 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 2 + e^{- 3 x} \) 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.
- \[ \frac{d}{d x} \left(2 + e^{- 3 x}\right) + 3 e^{- 3 x} = 0 \]Substituting y = 2 + 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 omits the constant term –6 from the differential equation. The substitution should yield y' + 3y – 6 = 0, not just y' + 3e^{-3x} = 0.qwen3.6:27b-mlx: fail (error) — The verification step is algebraically incorrect. Substituting y = 2 + e^{-3x} into 3y + y' - 6 yields 3(2 + e^{-3x}) + (-3e^{-3x}) - 6 = 0, but the solution claims the derivative term is +3e^{-3x} (implying y' = 3e^{-3x}), which is false since y' = -3e^{-3x}. The equation shown in line 3 does not match the substitution performed.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The verification step is algebraically incorrect. Substituting y = 2 + e^{-3x} into 3y + y' - 6 yields 3(2 + e^{-3x}) + (-3e^{-3x}) - 6 = 0, but the solution claims the derivative term is +3e^{-3x} (implying y' = 3e^{-3x}), which is false since y' = -3e^{-3x}. The equation shown in line 3 does not match the substitution performed.gpt-oss:20b: fail (error) 2026-10-18 — The verification step omits the constant term –6 from the differential equation. The substitution should yield y' + 3y – 6 = 0, not just y' + 3e^{-3x} = 0.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The verification step is algebraically incorrect. Substituting y = 2 + e^{-3x} into 3y + y' - 6 yields 3(2 + e^{-3x}) + (-3e^{-3x}) - 6 = 0, but the solution claims the derivative term alone plus 3*exp(-3x) equals 0, omitting the constant terms from 3y and -6. The equation shown does not match the substitution into the original ODE.gpt-oss:20b: fail (error) 2026-10-16 — The verification step omits the 3y and –6 terms; it only checks y′+3e^{–3x}=0, which is not the full differential equation. The student would learn an incorrect method for checking solutions.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The verification step is algebraically incorrect. Substituting y = 2 + e^{-3x} into 3y + y' - 6 yields 3(2 + e^{-3x}) + (-3e^{-3x}) - 6 = 0, but the solution's equation line shows 3*exp(-3*x) instead of 3*(2 + exp(-3*x)), implying the constant term 2 was dropped or mishandled in the substitution display.gpt-oss:20b: pass 2026-10-15
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.