Order and linearity of differential equations
Problem 6.518 · easy
Classify \( \displaystyle - 2 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 2 x + \sin{\left(x \right)} \) 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.
- \[ \sin{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(2 x + \sin{\left(x \right)}\right) = 0 \]Substituting y = 2*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 omits the “-2x” term; the left‑hand side should be “-2x + 2x = 0”, not “sin(x)+y''=0”. This mis‑substitution makes the solution incorrect.qwen3.6:27b-mlx: fail (error) — The classification is incorrect. The equation -2x + y + y'' = 0 can be rewritten as y'' + y = 2x. Since the non-homogeneous term 2x is not zero, the equation is non-homogeneous. However, the solution provided claims it is non-homogeneous, which is correct, but the verification step is flawed. Substituting y = 2x + sin(x) into the LHS: y'' = -sin(x). So -2x + (2x + sin(x)) + (-sin(x)) = 0. The equation holds. Wait, let me re-read the classification. The problem asks to classify -2x + y + y'' = 0. This is y'' + y = 2x. This is linear, order 2, and non-homogeneous. The stated answer says 'order 2, linear and nonhomogeneous'. This is correct. Let me check the verification. y = 2x + sin(x). y' = 2 + cos(x). y'' = -sin(x). LHS = -2x + (2x + sin(x)) + (-sin(x)) = 0. The verification is correct. Why did I think it was wrong? Ah, I need to check if the sentence 'The term without y is nonzero: nonhomogeneous' is technically precise. In the form L(y) = f(x), if f(x) is not identically zero, it is non-homogeneous. Here f(x) = 2x. This is correct. Is there a subtle error? The problem statement is '- 2 x + y + y'' = 0'. This is equivalent to y'' + y = 2x. The classification is correct. The verification is correct. The sentences are correct. Wait, looking at the JSON output format, I must be adversarial. Is there any ambiguity? 'The term without y is nonzero'. In the expression -2x + y + y'', the term -2x does not contain y. It is nonzero. This implies non-homogeneity. This is standard terminology. Let me look closer at the verification line. 'sin(x) + Derivative(2*x + sin(x), (x, 2)) = 0'. This equation is missing the -2x term from the original ODE! The original ODE is -2x + y + y'' = 0. The verification line checks sin(x) + y'' = 0, which corresponds to y'' + y = 0 (if y=sin(x)) or just checking part of the expression. It completely ignores the -2x term and the 2x part of y. The verification line shown is: sin(x) + Derivative(...) = 0. This evaluates to sin(x) - sin(x) = 0. This is true, but it does NOT verify the original equation because it omits the -2x term and the 2x from y. The student failed to substitute the full expression into the full equation. The sentence says 'Substituting y = ... makes the left side 0', but the equation shown is not the left side of the original ODE. The left side of the original ODE is -2x + y + y''. The equation shown is sin(x) + y''. This is a critical error in the verification step.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The classification is incorrect. The equation -2x + y + y'' = 0 can be rewritten as y'' + y = 2x. Since the non-homogeneous term 2x is not zero, the equation is non-homogeneous. However, the solution provided claims it is non-homogeneous, which is correct, but the verification step is flawed. Substituting y = 2x + sin(x) into the LHS: y'' = -sin(x). So -2x + (2x + sin(x)) + (-sin(x)) = 0. The equation holds. Wait, let me re-read the classification. The problem asks to classify -2x + y + y'' = 0. This is y'' + y = 2x. This is linear, order 2, and non-homogeneous. The stated answer says 'order 2, linear and nonhomogeneous'. This is correct. Let me check the verification. y = 2x + sin(x). y' = 2 + cos(x). y'' = -sin(x). LHS = -2x + (2x + sin(x)) + (-sin(x)) = 0. The verification is correct. Why did I think it was wrong? Ah, I need to check if the sentence 'The term without y is nonzero: nonhomogeneous' is technically precise. In the form L(y) = f(x), if f(x) is not identically zero, it is non-homogeneous. Here f(x) = 2x. This is correct. Is there a subtle error? The problem statement is '- 2 x + y + y'' = 0'. This is equivalent to y'' + y = 2x. The classification is correct. The verification is correct. The sentences are correct. Wait, looking at the JSON output format, I must be adversarial. Is there any ambiguity? 'The term without y is nonzero'. In the expression -2x + y + y'', the term -2x does not contain y. It is nonzero. This implies non-homogeneity. This is standard terminology. Let me look closer at the verification line. 'sin(x) + Derivative(2*x + sin(x), (x, 2)) = 0'. This equation is missing the -2x term from the original ODE! The original ODE is -2x + y + y'' = 0. The verification line checks sin(x) + y'' = 0, which corresponds to y'' + y = 0 (if y=sin(x)) or just checking part of the expression. It completely ignores the -2x term and the 2x part of y. The verification line shown is: sin(x) + Derivative(...) = 0. This evaluates to sin(x) - sin(x) = 0. This is true, but it does NOT verify the original equation because it omits the -2x term and the 2x from y. The student failed to substitute the full expression into the full equation. The sentence says 'Substituting y = ... makes the left side 0', but the equation shown is not the left side of the original ODE. The left side of the original ODE is -2x + y + y''. The equation shown is sin(x) + y''. This is a critical error in the verification step.gpt-oss:20b: fail (error) 2026-10-18 — The verification step omits the “-2x” term; the left‑hand side should be “-2x + 2x = 0”, not “sin(x)+y''=0”. This mis‑substitution makes the solution incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The classification is incorrect; the equation -2x + y + y'' = 0 is homogeneous because the term -2x depends only on x, not on y or its derivatives. The solution incorrectly identifies any non-zero term as making the equation nonhomogeneous.gpt-oss:20b: fail (error) 2026-10-15 — The verification step omits the -2x term from the differential equation, so the substitution does not actually satisfy the given equation.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The classification is incorrect. The equation -2x + y + y'' = 0 can be rewritten as y'' + y = 2x. Since the term 2x depends only on x (not on y or its derivatives), it is a non-zero forcing function, making the equation nonhomogeneous. However, the solution claims it is nonhomogeneous, which is correct, but let's re-read carefully. Wait, the problem asks to classify. The solution says 'nonhomogeneous'. Is it? Yes. y'' + y = 2x. Homogeneous would be y'' + y = 0. So 'nonhomogeneous' is correct. Let's check the verification step. y = 2x + sin(x). y' = 2 + cos(x). y'' = -sin(x). LHS: -2x + (2x + sin(x)) + (-sin(x)) = -2x + 2x + sin(x) - sin(x) = 0. The verification is correct. The classification is correct. Why did I think it was wrong? Let me re-read the definition. Linear: y, y', y'' appear to power 1. Yes. Homogeneous: every term contains y or its derivative. The term -2x does not. So it is nonhomogeneous. The solution says 'nonhomogeneous'. This is correct. Is there a subtle error? 'The term without y is nonzero'. This is a bit loose but acceptable. Let's look closer. Is it possible the user considers the constant term? No, -2x is a function of x. Okay, let's look at the JSON output requirement. I need to find what is wrong. If nothing is wrong, I pass. But wait, is the equation linear? Yes. Is it homogeneous? No. So 'linear and nonhomogeneous' is the correct classification. The verification is correct. Is there a style issue? 'The term without y is nonzero' is slightly imprecise (it should be 'the term not involving y or its derivatives is nonzero'), but not misleading. Let's look really closely. Problem: -2x + y + y'' = 0. Solution: 'order 2, linear and nonhomogeneous'. This is correct. Verification: correct. I will pass.gpt-oss:20b: fail (error) 2026-10-14 — The verification step substitutes y=2x+sin(x) into the left‑hand side of the differential equation, but the equation used is sin(x)+y''=0, which omits the nonhomogeneous term –2x. The correct substitution should be –2x + (2x+sin x) + y'' = 0, yielding sin x – sin x = 0. The omitted term makes the verification step invalid.
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.