Home›Calculus 2›Order and linearity of differential equations›Problem 6.418

Order and linearity of differential equations

Problem 6.418 · easy

Classify \( \displaystyle x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = - x + \sin{\left(x \right)} \) is a solution.

The highest derivative is y′′, so the order is 2.
Reviewed
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.
Reviewed
\[ \sin{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(- x + \sin{\left(x \right)}\right) = 0 \]
Substituting y = -x + sin(x) makes the left side 0.✓ Proved

Answer \( \text{order 2, linear and nonhomogeneous} \)

Lines: 1 proved, 2 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass
qwen3.6:27b-mlx: pass — The classification of the ODE as second-order, linear, and nonhomogeneous is correct. The verification step correctly substitutes the proposed solution into the differential equation.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The classification of the ODE as second-order, linear, and nonhomogeneous is correct. The verification step correctly substitutes the proposed solution into the differential equation.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-14 — The classification of the differential equation is correct (order 2, linear, nonhomogeneous). The verification step correctly substitutes the proposed solution into the differential equation and confirms it satisfies the equation.
gpt-oss:20b: fail (error) 2026-10-14 — The verification step substitutes the proposed solution into an incorrect expression; it omits the terms x and y from the original differential equation, so the conclusion that the left side is zero is unfounded.
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly classifies the differential equation as nonhomogeneous. The term 'x' is a function of the independent variable only, not a term involving the dependent variable y, so it does not affect homogeneity. The equation is homogeneous because every term (including the source term 0 on the RHS) is linear in y and its derivatives, or zero. Specifically, L[y] = y'' + y, and L[0] = 0, satisfying the definition of homogeneity. The presence of 'x' on the LHS makes it a non-homogeneous equation ONLY if the equation were written as y'' + y = f(x) with f(x) != 0. Here, the equation is y'' + y + x = 0, which can be rewritten as y'' + y = -x. In standard terminology, a linear ODE L(y) = g(x) is homogeneous if g(x) = 0. Here g(x) = -x, which is not zero. Wait, let me re-evaluate standard definitions carefully. Standard Definition: A linear ODE is homogeneous if it can be written as L(y) = 0, where L is a linear operator. If it is L(y) = g(x) with g(x) not identically zero, it is non-homogeneous. The given equation is y'' + y + x = 0. This can be written as y'' + y = -x. Here, the linear operator is L(y) = y'' + y. The right hand side is g(x) = -x. Since g(x) is not identically zero, the equation is NON-HOMOGENEOUS. Let me double check this. Many textbooks define a linear ODE a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = g(x). If g(x) = 0, it is homogeneous. If g(x) != 0, it is non-homogeneous. In the problem: y'' + y + x = 0 => y'' + y = -x. So g(x) = -x. Thus, it IS non-homogeneous. So the classification 'nonhomogeneous' is CORRECT. Let me re-read the solution's reasoning. 'The term without y is nonzero: nonhomogeneous.' This is slightly imprecise language but leads to the correct conclusion. The term 'x' does not contain y. It is part of the 'forcing function' or 'source term'. Let's check the verification step. y = -x + sin(x) y' = -1 + cos(x) y'' = -sin(x) LHS: y'' + y + x = -sin(x) + (-x + sin(x)) + x = -sin(x) - x + sin(x) + x = 0. So y is a solution. The solution says: 1. Order 2. Correct. 2. Linear. Correct. 3. Nonhomogeneous. Correct. Is there any error? 'The term without y is nonzero: nonhomogeneous.' This is the standard heuristic. If there is a term that does not depend on y or its derivatives, and it is not zero, the equation is non-homogeneous. Wait, is it possible the question implies the form L(y) = 0? No, the standard classification is based on the source term. Let's look closer at the prompt's provided 'Stated answer': 'order 2, linear and nonhomogeneous'. The solution matches this. Is there a subtle error? Some contexts define homogeneity differently? No, in standard ODE courses, y'' + y = -x is non-homogeneous. Let's check the sentence: 'The term without y is nonzero: nonhomogeneous.' This is technically correct reasoning for linear ODEs. However, look at the equation line: `sin(x) + Derivative(-x + sin(x), (x, 2)) = 0` This substitutes y = -x + sin(x) into `y'' + y`? No, the original equation is `y'' + y + x = 0`. The substitution in line 3 is: `sin(x)` (which is y''? No. y'' = -sin(x). So -sin(x)?) Let's calculate y'' for y = -x + sin(x). y' = -1 + cos(x) y'' = -sin(x) So y'' + y + x = -sin(x) + (-x + sin(x)) + x = 0. The equation in line 3 is: `sin(x) + Derivative(-x + sin(x), (x, 2)) = 0` Derivative(-x + sin(x), (x, 2)) is y'' = -sin(x). So the LHS is `sin(x) + (-sin(x)) = 0`. This equation `sin(x) + y'' = 0` is NOT the original ODE. The original ODE is `y'' + y + x = 0`. The verification step omitted the `+ x` term and the `y` term (replaced by sin(x)? No, y is -x+sin(x)). Let's trace the substitution in line 3 carefully. The code checks: `sin(x) + Derivative(...) = 0`. This corresponds to checking if `y'' + sin(x) = 0`? Or `y'' + y_part = 0`? If the student meant to substitute into `y'' + y + x = 0`, they should have written: `Derivative(y, (x,2)) + y + x = 0`. Instead, they wrote `sin(x) + y'' = 0`. Where did `sin(x)` come from? It looks like they substituted `y` with `sin(x)`? But y is `-x + sin(x)`. Or did they substitute `y` with `sin(x)` and ignore `-x`? If they ignored `-x`, they are verifying a different equation. Actually, looking at the expression `sin(x) + Derivative(-x + sin(x), (x, 2))`: `Derivative(-x + sin(x), (x, 2))` is `-sin(x)`. So `sin(x) - sin(x) = 0`. This verifies that `y'' + sin(x) = 0`. But the original equation is `y'' + y + x = 0`. Substituting `y = -x + sin(x)` into `y'' + y + x`: `-sin(x) + (-x + sin(x)) + x = 0`. The verification step in line 3 is **incorrect** because it does not substitute the full expression into the full equation. It seems to verify `y'' + sin(x) = 0` or some mangled version. It fails to include the `+ x` term and the `-x` part of `y`. Therefore, the verification is flawed/wrong. Verdict: fail. Severity: error. Notes: The verification step substitutes an incomplete expression into an incomplete equation, failing to actually verify the given solution against the given ODE.
gpt-oss:20b: fail (error) 2026-10-14 — The verification step omits the non‑zero term x from the differential equation; the left‑hand side should be x + y + y'' and only then can one check that y = -x + sin(x) satisfies it.

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.