Order and linearity of differential equations
Problem 6.423 · easy
Classify \( \displaystyle - 3 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 3 x + \sin{\left(x \right)} \) is a solution.
- The highest derivative is y′′, so the order is 2.
- 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.
- \[ \sin{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(3 x + \sin{\left(x \right)}\right) = 0 \]Substituting y = 3*x + sin(x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and nonhomogeneous} \)
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 y=3x+sin(x) but the displayed equation omits the -3x term, so it does not correctly check that the left‑hand side equals zero. The student would learn an incorrect substitution test.qwen3.6:27b-mlx: fail (error) — The classification is incorrect. The equation y'' + y = 3x is nonhomogeneous because of the 3x term on the RHS, but the solution verification step is flawed: substituting y = 3x + sin(x) into the LHS yields 3x + sin(x) + (-sin(x)) = 3x, which equals the RHS 3x, not 0. The equation line claims the LHS equals 0, which is false for this ODE (it should equal 3x). Furthermore, the problem asks to verify it is a solution, which implies checking LHS = RHS, but the equation shown checks LHS = 0, implying the ODE was treated as homogeneous in the verification step despite being classified as nonhomogeneous.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The classification is incorrect. The equation y'' + y = 3x is nonhomogeneous because of the 3x term on the RHS, but the solution verification step is flawed: substituting y = 3x + sin(x) into the LHS yields 3x + sin(x) + (-sin(x)) = 3x, which equals the RHS 3x, not 0. The equation line claims the LHS equals 0, which is false for this ODE (it should equal 3x). Furthermore, the problem asks to verify it is a solution, which implies checking LHS = RHS, but the equation shown checks LHS = 0, implying the ODE was treated as homogeneous in the verification step despite being classified as nonhomogeneous.gpt-oss:20b: fail (error) 2026-10-17 — The verification step substitutes y=3x+sin(x) but the displayed equation omits the -3x term, so it does not correctly check that the left‑hand side equals zero. The student would learn an incorrect substitution test.qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The classification is incorrect. The equation y'' + y = 3x is linear and non-homogeneous, but the solution claims it is homogeneous in the text (orgpt-oss:20b: fail (error) 2026-10-14 — The verification step omits the non‑homogeneous term –3x; the left side should be -3x + (3x+sin x) + (-sin x) = 0, not sin(x)+y''=0.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The classification is incorrect. The equation -3x + y + y'' = 0 can be rewritten as y'' + y = 3x. Since the right-hand side is not zero, it is nonhomogeneous. However, the solution verification step claims that substituting y = 3x + sin(x) makes the left side 0. Let's check: y = 3x + sin(x), y'' = -sin(x). LHS = -3x + (3x + sin(x)) + (-sin(x)) = 0. So y is indeed a solution. Wait, the classification says 'nonhomogeneous'. Is it? Yes, because of the -3x term. The stated answer says 'nonhomogeneous'. The solution text says 'The term without y is nonzero: nonhomogeneous.' This is correct reasoning for nonhomogeneity. Why did I think it was wrong? Let me re-read carefully. Problem: -3x + y + y'' = 0. Standard form: y'' + y = 3x. Homogeneous part: y'' + y = 0. Since 3x != 0, it is nonhomogeneous. Order 2: Correct. Linear: Correct. Nonhomogeneous: Correct. Let's check the verification step again. y = 3x + sin(x) y' = 3 + cos(x) y'' = -sin(x) Substitute into -3x + y + y'' = 0: -3x + (3x + sin(x)) + (-sin(x)) = -3x + 3x + sin(x) - sin(x) = 0. The equation holds. So the classification is correct. The verification is correct. The reasoning is correct. Wait, look at line 3: `[equation, unchecked] sin(x) + Derivative(3*x + sin(x), (x, 2)) = 0`. The original equation is `-3x + y + y'' = 0`. The substitution should result in `-3x + y + y'' = 0`. Line 3 shows `sin(x) + Derivative(...) = 0`. It is missing the `-3x` term and the `3x` part of `y`. `y = 3x + sin(x)`. `y'' = -sin(x)`. `-3x + y + y'' = -3x + 3x + sin(x) - sin(x) = 0`. Line 3 writes `sin(x) + Derivative(...) = 0`. `Derivative(3*x + sin(x), (x, 2))` is `-sin(x)`. So line 3 asserts `sin(x) - sin(x) = 0`, which is `0=0`. But it does not show the full substitution into the original equation `-3x + y + y''`. It seems to have implicitly cancelled `-3x` and `3x` from `y` without showing it, or it is checking a different equation. Actually, if you substitute `y = 3x + sin(x)` into `-3x + y + y''`, you get `-3x + 3x + sin(x) - sin(x)`. Line 3 only shows `sin(x) + y'' = 0`. This corresponds to the equation `y + y'' = 0` (if we ignore the 3x part of y?) No, `y` is `3x + sin(x)`. If the equation was `y + y'' = 0`, then `3x + sin(x) - sin(x) = 3x != 0`. The line `sin(x) + Derivative(...) = 0` is mathematically true (`sin(x) - sin(x) = 0`), but it does not represent the substitution into the given ODE `-3x + y + y'' = 0`. It represents the substitution into `y_part_from_sin + y'' = 0`? No. It looks like the solver substituted `y = sin(x)` into `y + y'' = 0`? Or perhaps they substituted `y = 3x + sin(x)` but forgot the `-3x` term in the equation and the `3x` term in `y`? If they forgot `-3x` in the equation and `3x` in `y`, they are checking `sin(x) + (-sin(x)) = 0`. This is a valid identity, but it doesn't verify the solution to the *given* problem because it ignores the non-homogeneous part `-3x` and the particular solution part `3x`. A student might think the `-3x` term doesn't matter or that `3x` isn't part of the solution check. This is misleading/error in the verification step description. The sentence in line 2 says 'nonhomogeneous'. The verification step fails to demonstrate how the nonhomogeneous term `-3x` is cancelled by the `3x` in `y`. It just shows the homogeneous part works for the sine component. This is an incomplete verification that could lead a student to believe the `-3x` term is irrelevant or that `y=3x` is not needed for the check. Therefore, the verification step is flawed/misleading.gpt-oss:20b: fail (error) 2026-10-14 — The verification step omits the non‑homogeneous term –3x, so the substituted expression does not actually satisfy the original differential equation.
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.