Home›Calculus 3›Variation of parameters›Problem 6.668

Variation of parameters

Problem 6.668 · hard

Use variation of parameters to find a particular solution of \( \displaystyle 4 y + y'' = \frac{1}{\cos{\left(2 x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(2 x \right)} \) and \( \displaystyle y_2 = \sin{\left(2 x \right)} \) solve the homogeneous equation.

\[ - \sin{\left(2 x \right)} \frac{d}{d x} \cos{\left(2 x \right)} + \cos{\left(2 x \right)} \frac{d}{d x} \sin{\left(2 x \right)} = 2 \]
The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
\[ \frac{d}{d x} \frac{\ln{\left(\cos{\left(2 x \right)} \right)}}{4} = - \frac{\tan{\left(2 x \right)}}{2} \]
u₁′ = −y₂g/W; u₁ = log(cos(2*x))/4.✓ Proved
\[ \frac{d}{d x} \frac{x}{2} = \frac{1}{2} \]
u₂′ = y₁g/W; u₂ = x/2.✓ Proved
y_p = u₁y₁ + u₂y₂: x*sin(2*x)/2 + log(cos(2*x))*cos(2*x)/4.
Reviewed
\[ 2 x \sin{\left(2 x \right)} + \ln{\left(\cos{\left(2 x \right)} \right)} \cos{\left(2 x \right)} + \frac{d^{2}}{d x^{2}} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\ln{\left(\cos{\left(2 x \right)} \right)} \cos{\left(2 x \right)}}{4}\right) = \frac{1}{\cos{\left(2 x \right)}} \]
It satisfies the equation.✓ Proved

Answer \( y_p = \frac{x \sin{\left(2 x \right)}}{2} + \frac{\ln{\left(\cos{\left(2 x \right)} \right)} \cos{\left(2 x \right)}}{4} \)

Lines: 4 proved, 1 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	✓ 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
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 — The solution correctly applies the method of variation of parameters, including the necessary step of dividing the non-homogeneous term by the leading coefficient (4) to identify g(x). The Wronskian calculation and subsequent integrations are correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly applies the method of variation of parameters, including the necessary step of dividing the non-homogeneous term by the leading coefficient (4) to identify g(x). The Wronskian calculation and subsequent integrations are correct.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution fails to normalize the differential equation to standard form (y'' + P(x)y' + Q(x)y = g(x)) before applying variation of parameters. The forcing term g(x) should be 1/(4 cos(2x)), not 1/cos(2x), leading to incorrect integrals for u1 and u2.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution fails to normalize the differential equation to standard form y'' + p(x)y' + q(x)y = g(x) before applying variation of parameters. The given equation is 4y + y'' = 1/cos(2x), so the forcing term for the method should be g(x) = 1/(4 cos(2x)), not 1/cos(2x). Consequently, the integrals for u1' and u2' are off by a factor of 4, leading to an incorrect particular solution.
gpt-oss:20b: pass 2026-10-16

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 2026-10-19 with SymPy 1.14.0.