Home›Calculus 2›Order and linearity of differential equations›Problem 6.519

Order and linearity of differential equations

Problem 6.519 · easy

Classify \( \displaystyle - x + y y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \sqrt{x^{2} + 3} \) is a solution.

The highest derivative is y′, so the order is 1.
It is nonlinear: y appears squared or multiplied by its own derivative.
\[ - x + \sqrt{x^{2} + 3} \frac{d}{d x} \sqrt{x^{2} + 3} = 0 \]
Substituting y = sqrt(x**2 + 3) makes the left side 0.✓ Proved

Answer \( \text{order 1, nonlinear} \)

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: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly classifies the ODE as nonlinear. The equation -x + y y' = 0 is quasilinear (linear in the highest derivative y') and is oft

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-18 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly classifies the ODE as nonlinear. The equation -x + y y' = 0 is quasilinear (linear in the highest derivative y') and is oft
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The solution incorrectly classifies the ODE as nonlinear. The equation -x + y y' = 0 can be rewritten as y' - (1/y)x = 0, which is not linear in y, but the standard definition of linearity for first-order ODEs requires the equation to be of the form y' + P(x)y = Q(x). The term y y' makes it nonlinear in the dependent variable y? Wait. Let's re-evaluate. A first-order ODE is linear if it can be written as y' + P(x)y = Q(x). Here we have y y' = x, or y' = x/y. This is not linear because of the 1/y term (or y y' term). So the classification 'nonlinear' is actually correct. Why did I think it was an error? Let me re-read the definition. Linear ODE: L(y) = g(x) where L is a linear operator. L(y) = y' + p(x)y. Here we have y y'. This is not linear. So the classification is correct. Is there any other issue? The verification step is marked as proved. The sentence 'y appears squared or multiplied by its own derivative' is a bit loose but generally acceptable as a heuristic for nonlinearity (specifically the y y' term). However, looking closer at the prompt's provided answer: 'order 1, nonlinear'. The solution says 'nonlinear'. This is correct. Is there a subtle error? Maybe the problem implies checking if it's homogeneous? The problem asks 'whether it is linear and, if so, whether it is homogeneous'. Since it's nonlinear, the 'if so' part is skipped. The solution does this correctly. Let me look for a different error. Ah, look at the verification. y = sqrt(x^2 + 3). y' = x / sqrt(x^2 + 3). LHS = -x + sqrt(x^2+3) * (x / sqrt(x^2+3)) = -x + x = 0. This is correct. Is the classification wrong? Some definitions might consider y y' = x as a Bernoulli equation or something else, but it is definitely nonlinear. Wait, is it possible the question considers it linear in x? No, standard classification is in y. Let me check if I missed something about 'homogeneous'. A linear ODE is homogeneous if Q(x) = 0. Since it's nonlinear, this doesn't apply. The solution seems correct. Why would it fail? Let me re-read the sentence: 'y appears squared or multiplied by its own derivative.' In the equation -x + y y' = 0, y is not squared. It is multiplied by its derivative. The sentence says 'y appears squared OR multiplied...'. This is a true statement about the structure (the second part of the OR is true). It's a bit imprecise but not false. Is there a convention where y y' = 0 is considered linear? No. Let me check the provided answer again. 'order 1, nonlinear'. The solution matches. Is there a hidden trap? Maybe the domain? No. Let's look at the JSON output requirements. If it's sound, pass. If I can't find an error, I should pass. But I am an adversarial reviewer. Let me look really closely. 'y appears squared'. It does not. 'or multiplied by its own derivative'. It is. So the condition for nonlinearity is met. The classification is correct. The verification is correct. I will mark it as pass. Wait, let me double check the definition of linear. An ODE is linear if it is linear in y and y'. The term y y' is a product of y and y', so it is nonlinear. Correct. Okay, I will pass it.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly classifies the differential equation as nonlinear. The equation -x + yy' = 0 is linear in y (specifically, it can be writte
gpt-oss:20b: pass 2026-10-14

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.