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Undetermined coefficients

Problem 6.248 · hard

Find the general solution of \( \displaystyle y'' + 3y' + 2y = 5 x + 6 \) by the method of undetermined coefficients.

\[ r^{2} + 3 r + 2 = \left(r + 1\right) \left(r + 2\right) \]
The characteristic equation has roots -2 and -1.✓ Proved
So y_h = C₁e^(-2x) + C₂e^(-1x). Guess y_p = A*x + B.
Reviewed
\[ 5 x + 3 \frac{d}{d x} \left(\frac{5 x}{2} - \frac{3}{4}\right) + \frac{d^{2}}{d x^{2}} \left(\frac{5 x}{2} - \frac{3}{4}\right) - \frac{3}{2} = 5 x + 6 \]
Matching coefficients gives y_p = 5*x/2 - 3/4; it satisfies the equation.✓ Proved

Answer \( y = C_{1} e^{- 2 x} + C_{2} e^{- x} + \frac{5 x}{2} - \frac{3}{4} \)

Lines: 2 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.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
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	differs from SymPy's dsolve only by a solution of the homogeneous equation

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the homogeneous solution and the form of the particular solution. The coefficients are calculated correctly, and the final general solution is accurate.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the homogeneous solution and the form of the particular solution. The coefficients are calculated correctly, and the final general solution is accurate.
gpt-oss:20b: pass 2026-10-05
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The particular solution y_p = 5x/2 - 3/4 is incorrect. Substituting this into the ODE yields 5x + 3(5/2) + 2(-3/4) = 5x + 15/2 - 3/2 = 5x + 6, which actually works. Wait, let me re-check. y_p = Ax + B. y_p' = A, y_p'' = 0. ODE: 0 + 3A + 2(Ax + B) = 5x + 6 => 2Ax + (3A + 2B) = 5x + 6. 2A = 5 => A = 5/2. 3(5/2) + 2B = 6 => 15/2 + 2B = 12/2 => 2B = -3/2 => B = -3/4. The solution is correct. Why did I think it was wrong? Let me re-read the prompt's solution line 3. It says 'Matching coefficients gives y_p = 5*x/2 - 3/4; it satisfies the equation.' The equation shown is '5*x + 3*Derivative(5*x/2 - 3/4, x) + Derivative(5*x/2 - 3/4, (x, 2)) - 3/2 = 5*x + 6'. This equation is nonsensical. It seems to be trying to verify the solution but has a stray '- 3/2' and the structure is garbled. The verification step is presented as an equation that is 'unchecked' by the CAS but presumably correct? No, the prompt says 'Each line... is either an EQUATION, which a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. Line 3 is marked [equation, unchecked]. This is contradictory. Usually 'unchecked' means the CAS didn't check it, or it's not marked 'proved'. The prompt says 'do not re-check the algebra of a line marked proved'. Line 3 is NOT marked proved. It is marked 'unchecked'. So I should check it? Or is 'unchecked' a label for the type? The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Line 3 is an EQUATION. The text says 'its result is shown'. The result shown is '5*x + 3*Derivative(...) ... = 5*x + 6'. This equality is FALSE as written because of the '- 3/2' term on the LHS. 5x + 3(5/2) + 0 - 3/2 = 5x + 15/2 - 3/2 = 5x + 6. So the equation IS true. 5x + 6 = 5x + 6. Okay, so the equation is correct. The sentence in line 2 says 'Guess y_p = A*x + B'. This is correct. The sentence in line 1 says 'The characteristic equation has roots -2 and -1'. This is correct. The final answer is correct. Is there a subtle error? 'y_h = C₁e^(-2x) + C₂e^(-1x)'. Standard form is usually e^{-x} and e^{-2x}. Order doesn't matter. The method is undetermined coefficients. The guess is correct. The calculation is correct. The verification is correct. Why would it fail? Maybe the sentence 'So y_h = ...' implies that the homogeneous solution is derived directly from the roots, which is standard. Maybe the issue is that the solution doesn't explicitly state the general solution y = y_h + y_p? The stated answer is correct. The solution steps are correct. Let me look closer at line 3. 'Matching coefficients gives y_p = 5*x/2 - 3/4; it satisfies the equation.' This is a sentence embedded in an equation line? No, the line is marked [equation]. The text after '--' is likely a comment. The equation itself is '5*x + 3*Derivative(5*x/2 - 3/4, x) + Derivative(5*x/2 - 3/4, (x, 2)) - 3/2 = 5*x + 6'. As calculated, LHS = 5x + 7.5 + 0 - 1.5 = 5x + 6. RHS = 5x + 6. It holds. Is there a pedagogical issue? The student might be confused by the '- 3/2' term. It looks like a typo for the constant term B? No, B is -3/4. 2B = -3/2. Ah, the term 2y_p contributes 2(-3/4) = -3/2. The term 3y_p' contributes 3(5/2) = 15/2. The term y_p'' is 0. The term 5x is 5x. So the LHS is 5x + 15/2 - 3/2. The equation writes it as 5x + 3*y_p' + y_p'' - 3/2. This is weird notation. It mixes the function y_p with its derivatives and then subtracts 3/2? It should be 5x + 3*y_p' + 2*y_p = 5x + 6. The equation written is 5x + 3*y_p' + y_p'' - 3/2 = 5x + 6. This is not the standard way to write the verification. It's a specific numerical substitution. It is technically correct but very confusing. However, the prompt asks to find what is WRONG. Is it wrong? No. Is it misleading? Maybe. But usually 'error' is for mathematical mistakes. Let's look at the roots again. r^2 + 3r + 2 = 0. (r+1)(r+2)=0. r=-1, -2. Correct. y_h = C1 e^{-x} + C2 e^{-2x}. Correct. Guess y_p = Ax+B. Correct. A=5/2, B=-3/4. Correct. General solution y = C1 e^{-2x} + C2 e^{-x} + 5x/2 - 3/4. Correct. I will mark it pass. Wait, looking at line 1: 'r**2 + 3*r + 2 = (r + 1)*(r + 2)'. This is an identity. The comment says 'The characteristic equation has roots -2 and -1'. This is a sentence. Line 2 is a sentence. Line 3 is an equation. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Line 1 is marked [equation, unchecked]. Line 2 is [sentence]. Line 3 is [equation, unchecked]. The review should focus on the sentences. Sentence 1 (comment on line 1): 'The characteristic equation has roots -2 and -1.' Correct. Sentence 2: 'So y_h = C₁e^(-2x) + C₂e^(-1x). Guess y_p = A*x + B.' Correct. Sentence 3 (comment on line 3): 'Matching coefficients gives y_p = 5*x/2 - 3/4; it satisfies the equation.' Correct. Everything seems correct. I will output pass.
gpt-oss:20b: pass 2026-10-05

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/undetermined_coefficients, checked 2026-10-05 with SymPy 1.14.0.