Home›Calculus 2›Order and linearity of differential equations›Problem 6.419

Order and linearity of differential equations

Problem 6.419 · easy

Classify \( \displaystyle - x + y y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \sqrt{x^{2} + 1} \) 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.
\[ - x + \sqrt{x^{2} + 1} \frac{d}{d x} \sqrt{x^{2} + 1} = 0 \]
Substituting y = sqrt(x**2 + 1) 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: fail (error) — Sentence 2 incorrectly claims that y appears squared; the equation contains a product y·y′, which makes it nonlinear, but the description is misleading.
qwen3.6:27b-mlx: fail (error) — 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 form y' + P(x)y = Q(x). Here, the term y y' makes it nonlinear in the dependent variable y if we consider the standard linear form. Wait, let's re-evaluate. A first-order ODE is linear if it can be written as y' + p(x)y = q(x). The given equation is 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. Let me re-read the prompt's stated answer. Stated answer: 'order 1, nonlinear'. The solution says 'nonlinear'. Why did I think it was wrong? Let's check the definition again. Linear ODE: L(y) = f(x) where L is a linear operator. The operator d/dx is linear. The term y y' is not linear in y. So it is nonlinear. The solution is correct. Why did I flag it? Let me look closer at the solution text. 'y appears squared or multiplied by its own derivative'. In -x + y y' = 0, y is multiplied by its derivative. This is a valid reason for nonlinearity. The verification step is also correct. Is there a subtle error? The problem asks to classify whether it is homogeneous. The solution does not address homogeneity. The problem statement says: 'Classify ... its order, whether it is linear and, if so, whether it is homogeneous.' Since it is nonlinear, the 'if so' condition fails, so we don't need to check homogeneity for the linear case. However, usually 'homogeneous' can refer to homogeneous functions (M/N) or homogeneous linear ODEs. Given the context 'if so, whether it is homogeneous', it implies homogeneous linear. Since it's not linear, we stop. The solution provides the classification 'order 1, nonlinear'. This matches the stated answer. Is there any error? Let's check the verification. y = sqrt(x^2+1). y' = x/sqrt(x^2+1). y y' = sqrt(x^2+1) * x/sqrt(x^2+1) = x. So -x + x = 0. Correct. The classification is correct. The reasoning is correct. Why would this fail? Maybe the prompt implies I should find an error. Let me re-read carefully. 'Classify ... whether it is linear and, if so, whether it is homogeneous.' The solution states 'order 1, nonlinear'. It does not explicitly state 'not homogeneous' but the condition 'if so' makes it irrelevant. However, a complete answer might explicitly say 'not linear, so homogeneity is not applicable' or similar. But the stated answer is just 'order 1, nonlinear'. The solution matches. Is there a mistake in the sentence 'y appears squared'? It doesn't appear squared. It appears multiplied by its derivative. The sentence says 'y appears squared OR multiplied by its own derivative'. This is a disjunction. One part is true. So the sentence is logically true. Is it misleading? It might suggest y^2 is present, which it isn't. But it's not false. Let's look for a deeper error. Is the ODE actually linear? No. Is the solution correct? Yes. Wait, is it possible the question considers it homogeneous in the sense of homogeneous functions? y' = x/y. M = -x, N = y. M(tx, ty) = -tx = t(-x) = tM(x,y). N(tx, ty) = ty = tN(x,y). It is homogeneous of degree 1. But the question asks 'if so [linear], whether it is homogeneous'. This restricts the homogeneity check to the linear case. So we don't need to check it. The solution is correct. I will mark it pass. Wait, looking at similar adversarial tasks, often 'nonlinear' is the trap if it *is* linear. But y y' is definitely nonlinear. What if the variable was t? No, x and y. Okay, I will mark pass.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — 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 form y' + P(x)y = Q(x). Here, the term y y' makes it nonlinear in the dependent variable y if we consider the standard linear form. Wait, let's re-evaluate. A first-order ODE is linear if it can be written as y' + p(x)y = q(x). The given equation is 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. Let me re-read the prompt's stated answer. Stated answer: 'order 1, nonlinear'. The solution says 'nonlinear'. Why did I think it was wrong? Let's check the definition again. Linear ODE: L(y) = f(x) where L is a linear operator. The operator d/dx is linear. The term y y' is not linear in y. So it is nonlinear. The solution is correct. Why did I flag it? Let me look closer at the solution text. 'y appears squared or multiplied by its own derivative'. In -x + y y' = 0, y is multiplied by its derivative. This is a valid reason for nonlinearity. The verification step is also correct. Is there a subtle error? The problem asks to classify whether it is homogeneous. The solution does not address homogeneity. The problem statement says: 'Classify ... its order, whether it is linear and, if so, whether it is homogeneous.' Since it is nonlinear, the 'if so' condition fails, so we don't need to check homogeneity for the linear case. However, usually 'homogeneous' can refer to homogeneous functions (M/N) or homogeneous linear ODEs. Given the context 'if so, whether it is homogeneous', it implies homogeneous linear. Since it's not linear, we stop. The solution provides the classification 'order 1, nonlinear'. This matches the stated answer. Is there any error? Let's check the verification. y = sqrt(x^2+1). y' = x/sqrt(x^2+1). y y' = sqrt(x^2+1) * x/sqrt(x^2+1) = x. So -x + x = 0. Correct. The classification is correct. The reasoning is correct. Why would this fail? Maybe the prompt implies I should find an error. Let me re-read carefully. 'Classify ... whether it is linear and, if so, whether it is homogeneous.' The solution states 'order 1, nonlinear'. It does not explicitly state 'not homogeneous' but the condition 'if so' makes it irrelevant. However, a complete answer might explicitly say 'not linear, so homogeneity is not applicable' or similar. But the stated answer is just 'order 1, nonlinear'. The solution matches. Is there a mistake in the sentence 'y appears squared'? It doesn't appear squared. It appears multiplied by its derivative. The sentence says 'y appears squared OR multiplied by its own derivative'. This is a disjunction. One part is true. So the sentence is logically true. Is it misleading? It might suggest y^2 is present, which it isn't. But it's not false. Let's look for a deeper error. Is the ODE actually linear? No. Is the solution correct? Yes. Wait, is it possible the question considers it homogeneous in the sense of homogeneous functions? y' = x/y. M = -x, N = y. M(tx, ty) = -tx = t(-x) = tM(x,y). N(tx, ty) = ty = tN(x,y). It is homogeneous of degree 1. But the question asks 'if so [linear], whether it is homogeneous'. This restricts the homogeneity check to the linear case. So we don't need to check it. The solution is correct. I will mark it pass. Wait, looking at similar adversarial tasks, often 'nonlinear' is the trap if it *is* linear. But y y' is definitely nonlinear. What if the variable was t? No, x and y. Okay, I will mark pass.
gpt-oss:20b: fail (error) 2026-10-17 — Sentence 2 incorrectly claims that y appears squared; the equation contains a product y·y′, which makes it nonlinear, but the description is misleading.
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly classifies the ODE as nonlinear. The equation -x + yy' = 0 is linear in y (it can be written as y' + (0)y = x/y, wait, no. Standard form for linear is y' + P(x)y = Q(x). Here we have y y' = x. This is not linear in y because of the y*y' term? Actually, let's look closer. A first-order ODE is linear if it can be written as y' + p(x)y = q(x). The given equation is y y' = x. This cannot be put in that form without dividing by y, which makes it singular at y=0, but more importantly, the term y y' makes it non-linear in the dependent variable y? No, wait. The standard definition of a linear ODE is that the dependent variable and its derivatives appear to the first power and are not multiplied together. Here, y and y' are multiplied. So it IS nonlinear. Wait, let me re-read the standard definition carefully. Definition: An ODE is linear if it is linear in the dependent variable y and all its derivatives. This means terms like y^2, (y')^2, y*y', sin(y) are forbidden. The term y*y' is a product of the dependent variable and its derivative. Therefore, the ODE is indeed nonlinear. So why did I think it might be linear? Sometimes students confuse 'linear' with 'homogeneous' or other properties. Or they might think of it as exact or separable. But strictly speaking, y y' = x is nonlinear. Let me double check. Is there any interpretation where it is linear? No. So the classification 'nonlinear' is correct. Let me re-read the prompt's stated answer: 'order 1, nonlinear'. The solution says: 'It is nonlinear: y appears squared or multiplied by its own derivative.' The reasoning 'y appears ... multiplied by its own derivative' is correct for the term y y'. So the verdict should be pass? Let me check the verification step. y = sqrt(x^2 + 1). y' = x / sqrt(x^2 + 1). LHS = -x + y y' = -x + sqrt(x^2+1) * (x / sqrt(x^2+1)) = -x + x = 0. This is correct. So the solution is correct. Wait, is there a nuance? Some texts define linear ODEs as those where the operator L[y] = 0 is linear. L[y] = y y' - x. L[ay] = ay (ay)' - x = a^2 y y' - x != a L[y]. So it is nonlinear. Okay, so the solution is correct. Why did I hesitate? I was worried about the 'homogeneous' part of the question. The question asks: 'Classify ... its order, whether it is linear and, if so, whether it is homogeneous.' The solution states it is nonlinear. It does not explicitly address the 'if so, whether it is homogeneous' part, but since it is not linear, the condition 'if so' is false, so one does not need to classify it as homogeneous or not in the linear sense. However, sometimes 'homogeneous' can refer to homogeneous functions (degree 0). y y' = x is homogeneous of degree 1? Let f(x,y) = x/y. No, dy/dx = x/y. f(tx, ty) = tx/ty = x/y = f(x,y). So it is a homogeneous ODE in the sense of homogeneous functions. But the question structure 'whether it is linear and, if so, whether it is homogeneous' strongly implies 'linear homogeneous'. Since it's not linear, we stop. The solution says 'nonlinear'. This is correct. Is there any error? 'y appears squared or multiplied by its own derivative'. In the equation -x + y y' = 0, y does not appear squared. It appears multiplied by its derivative. The sentence says 'y appears squared OR multiplied...'. This is a logical OR. Since the second part is true, the statement is true. It is slightly imprecise to say 'y appears squared' when it doesn't, but the 'or' saves it. It might be better to say 'y is multiplied by its derivative'. But is it an error? It's a bit sloppy. Let's look at the 'stated answer'. It just says 'order 1, nonlinear'. The solution matches the stated answer. I will mark it as pass. Wait, let me look really closely at the sentence: 'It is nonlinear: y appears squared or multiplied by its own derivative.' Does y appear squared? No. Does y appear multiplied by its own derivative? Yes. So the reason given is valid. Verdict: pass.
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The classification is incorrect; the equation -x + yy' = 0 is linear in y' and can be written as y' - (1/y)y = 0, but more importantly, standard definition of linearity for ODEs requires the dependent variable and its derivatives to appear to the first power and not multiplied together. Here y and y' are multiplied, making it nonlinear. However, the provided solution claims it is nonlinear but gives a vague reason ('y appears squared or multiplied by its own derivative'). The equation is indeed nonlinear because of the yy' term. Wait, let me re-read. The problem asks to classify. The solution says 'nonlinear'. This is correct. Why did I think it was linear? A linear ODE is of the form a_n(x)y^(n) + ... + a_1(x)y' + a_0(x)y = g(x). The term yy' makes it nonlinear. So the classification 'nonlinear' is correct. The reason 'y appears squared or multiplied by its own derivative' is slightly imprecise (y is not squared, but y*y' is a product of dependent variable and derivative), but the conclusion is correct. Let's look closer. Is it homogeneous? A linear ODE is homogeneous if g(x)=0. Since it's nonlinear, the term 'homogeneous' usually refers to the property f(tx, ty) = t^n f(x,y) or similar. The prompt asks 'if so [linear], whether it is homogeneous'. Since it's not linear, we don't need to check homogeneity in the linear sense. The solution stops at 'nonlinear'. This is correct. Let's re-evaluate the 'error'. The solution says 'y appears squared'. y does NOT appear squared in -x + yy' = 0. It appears as y * y'. Saying 'y appears squared' is factually false regarding the equation structure. This is a misleading statement about the equation's form, even if the final classification (nonlinear) is correct. A student might think y^2 is present. Also, the verification step is marked [equation, unchecked] but the text says 'Substituting ... makes the left side 0'. The algebra is correct. The main issue is the justification for nonlinearity. It claims 'y appears squared'. This is false. y is to the first power, but multiplied by y'. This is a 'misleading' or 'error' in the explanation. Given the strict adversarial role, claiming a term exists that doesn't is an error in the sentence.
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 2026-10-17 with SymPy 1.14.0.