Home›Calculus 3›Variation of parameters›Problem 6.511

Variation of parameters

Problem 6.511 · hard

Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{2 e^{x}}{x} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.

\[ - x e^{x} \frac{d}{d x} e^{x} + e^{x} \frac{d}{d x} x e^{x} = e^{2 x} \]
The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
\[ \frac{d}{d x} \left(- 2 x\right) = -2 \]
u₁′ = −y₂g/W; u₁ = -2*x.✓ Proved
\[ \frac{d}{d x} 2 \ln{\left(x \right)} = \frac{2}{x} \]
u₂′ = y₁g/W; u₂ = 2*log(x).✓ Proved
\[ 2 x e^{x} \ln{\left(x \right)} - 2 x e^{x} = 2 x \left(\ln{\left(x \right)} - 1\right) e^{x} \]
y_p = u₁y₁ + u₂y₂.✓ Proved
\[ 2 x \left(\ln{\left(x \right)} - 1\right) e^{x} - 2 \frac{d}{d x} 2 x \left(\ln{\left(x \right)} - 1\right) e^{x} + \frac{d^{2}}{d x^{2}} 2 x \left(\ln{\left(x \right)} - 1\right) e^{x} = \frac{2 e^{x}}{x} \]
It satisfies the equation.✓ Proved

Answer \( y_p = 2 x \left(\ln{\left(x \right)} - 1\right) e^{x} \)

Every line of this solution was proved by the computer algebra system SymPy. 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ 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	finite differences of the answer satisfy the equation at three points

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The solution incorrectly identifies the non-homogeneous term g(x) as 2e^x/x instead of 2/(x e^x) (since the standard form requires dividing by the leading coefficient 1, but the Wronskian calculation and subsequent integrals imply g(x) was taken as 2e^x/x while the formula u' = y g / W requires g to be the RHS of y'' + p y' + q y = g. Here the RHS is 2e^x/x, so g=2e^x/x. Wait, let's re-evaluate. The ODE is y'' - 2y' + y = 2e^x/x. So g(x) = 2e^x/x. W = e^{2x}. u1' = -y2 g / W = - (x e^x) (2e^x/x) / e^{2x} = -2. u2' = y1 g / W = (e^x) (2e^x/x) / e^{2x} = 2/x. The integrals u1 = -2x and u2 = 2ln|x| are correct. The final yp is correct. Why did I think it was wrong? Let me re-read the prompt's solution lines. Line 2 says u1' = -2. Line 3 says u2' = 2/x. These are correct. The final answer is correct. Is there a subtle error? The problem asks for *a* particular solution. The solution provides one. The steps are algebraically verified. However, usually ln(x) implies x>0. The problem doesn't specify domain. But more importantly, look at the Wronskian calculation in line 1. It calculates W = e^{2x}. This is correct. The formulas for u1' and u2' are correct. The integration is correct. The combination is correct. The verification is correct. Is there a style issue? "u1 = -2*x" is an antiderivative, technically u1 = -2x + C1. But for variation of parameters, we just need one particular solution, so setting constants to 0 is standard. The solution seems sound. Wait, let me check the sign in the Wronskian definition. W = y1 y2' - y2 y1'. y1 = e^x, y2 = x e^x. y1' = e^x, y2' = e^x + x e^x. W = e^x(e^x + x e^x) - x e^x(e^x) = e^{2x} + x e^{2x} - x e^{2x} = e^{2x}. Correct. u1' = -y2 g / W. u2' = y1 g / W. Correct. The solution is actually correct. Why would it fail? Maybe the prompt implies I should find an error. Let me look closer at the "Stated answer" vs the derived answer. Stated: 2 x (ln(x) - 1) e^x. Derived: 2 x (log(x) - 1) e^x. They match. Is there a missing absolute value in ln(x)? Usually, in calculus contexts, ln(x) is accepted for x>0. If the domain isn't specified, ln|x| is safer, but ln(x) is often used loosely. Is that an error? It's a minor style/domain issue. But wait, look at line 2: "u1 = -2*x". The derivative of -2x is -2. Correct. Line 3: "u2 = 2*log(x)". Derivative is 2/x. Correct. The solution is mathematically sound for x>0. I will mark it as pass.
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 fails to normalize the differential equation to standard form (coefficient of y'' is 1) before identifying g(x). The correct forcing te
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/variation_of_parameters, checked 2026-10-17 with SymPy 1.14.0.