Home›Calculus 3›Second-order linear equations›Problem 6.157

Second-order linear equations

Problem 6.157 · hard

Solve \( \displaystyle y'' - 1y' - 2y = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 0 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 2\right) \left(m + 1\right) = m^{2} - m - 2 \]
Roots: 2, -1.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \frac{\left(4 e^{3 x} + 8\right) e^{- x}}{3} - \frac{d}{d x} \frac{\left(- 2 e^{3 x} - 4\right) e^{- x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(- 2 e^{3 x} - 4\right) e^{- x}}{3} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -2 \]
y(0).✓ Proved
\[ 0 \]
y'(0).✓ Proved

Answer \( y = \frac{2 \left(- e^{3 x} - 2\right) e^{- x}}{3} \)

Lines: 4 proved, 3 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.

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Not checked	—	a sentence; read, not computed
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
7	✓ 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	sympy.checkodesol substitutes the solution back; both initial conditions match

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The characteristic equation is incorrectly stated as m² + bm + c = 0 without specifying b and c, leading to a sign error in the roots (should be m² - m - 2 = 0, giving roots 2 and -1, but the text implies a generic form that doesn't match the specific coefficients). More critically, the solution provided in the stated answer and checked in line 5 is 2*(-exp(3*x) - 2)*exp(-x)/3, which simplifies to -2/3 * exp(2x) + 4/3 * exp(-x). Let's check the roots: The characteristic equation m^2 - m - 2 = 0 has roots m = 2 and m = -1. The general solution is y = C1 e^(2x) + C2 e^(-x). y(0) = C1 + C2 = -2. y'(0) = 2C1 - C2 = 0 => C2 = 2C1. 3C1 = -2 => C1 = -2/3, C2 = 4/3. So y = -2/3 e^(2x) + 4/3 e^(-x). The stated answer is 2/3 * (-e^(2x) + 2e^(-x)) = -2/3 e^(2x) + 4/3 e^(-x). This matches. However, line 1 says 'the characteristic equation is m² + bm + c = 0'. For y'' - y' - 2y = 0, the equation is m² - m - 2 = 0. Line 2 shows (m-2)(m+1) which is m² - m - 2. The roots are 2 and -1. Line 3 says 'Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x)'. This is correct. The error is subtle: Line 1 is a generic statement, but it doesn't explicitly link b=-1, c=-2. Is this an error? It's a bit loose. Let's look closer. The prompt asks to find what is wrong. The solution steps are: 1. Setup char eq. 2. Factor/Roots. 3. General form. 4. Apply ICs. 5-7. Verification. The verification lines 5, 6, 7 check the *stated answer* against the ODE and ICs. They pass. The derivation in 1-4 is sketched. Is there a logical gap? Line 1 says 'Try y = e^(mx): the characteristic equation is m² + bm + c = 0'. This is a standard method description. It doesn't explicitly state b=-1, c=-2, but line 2 provides the specific polynomial. This is acceptable shorthand. Wait, look at line 2: '(m - 2)*(m + 1), m**2 - m - 2'. This confirms the polynomial. The roots are 2, -1. The general solution is C1 e^(2x) + C2 e^(-x). The constants are found to be -2/3 and 4/3. The stated answer is equivalent. The solution seems correct. Why would it fail? Let's re-read carefully. 'm² + bm + c = 0'. If the student writes this, they must identify b and c. The problem is y'' - 1y' - 2y = 0. So b = -1, c = -2. The sentence in line 1 is generic. It is not wrong, just generic. Is there a sign error in the roots? (m-2)(m+1) = m^2 - m - 2. Roots 2, -1. Correct. General solution form correct. Constants calculation is skipped but verified by CAS. The verification lines 5, 6, 7 confirm the final answer is correct. Is there a 'misleading' aspect? Line 1 might be considered 'style' if it's too vague, but it's not 'error'. However, often in these adversarial reviews, if the derivation is skipped and only the final answer is verified, it might be considered 'pass' if the verification is robust. But wait, look at line 5. It checks the ODE. Line 6 checks y(0). Line 7 checks y'(0). All proved. The sentences 1-4 describe the method. Is there a flaw in the method description? No. Is there a flaw in the setup? No. Maybe the 'error' is that the solution doesn't actually *show* the calculation of C1 and C2, just says 'Use y(0)... to find C1 and C2' and then jumps to verification. This is a 'style' issue or 'misleading' if it implies the verification is the derivation. But the prompt says 'Review worked calculus solutions'. If the work is missing, is it a fail? Usually, 'pass' is for sound solutions. A solution that skips the algebraic step of solving for constants but verifies the result is technically 'correct' but poor style. However, looking at similar examples, if the logic is sound and the result is verified, it often passes. Let's look for a harder error. Did I miss a sign? y'' - y' - 2y = 0. Char eq: r^2 - r - 2 = 0. (r-2)(r+1)=0. r=2, -1. y = c1 e^(2x) + c2 e^(-x). y(0) = c1+c2 = -2. y'(0) = 2c1 - c2 = 0. c2 = 2c1. 3c1 = -2 -> c1 = -2/3. c2 = 4/3. y = -2/3 e^(2x) + 4/3 e^(-x). Stated answer: 2*(-exp(3*x) - 2)*exp(-x)/3 = 2/3 * (-e^(2x) + 2e^(-x)) = -2/3 e^(2x) + 4/3 e^(-x). It matches. The solution is correct. Why would it be a fail? Perhaps the sentence 'Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x)' is considered incomplete because it doesn't specify that r1 and r2 are the roots found? No, that's implied. Is there a typo in the problem statement vs solution? No. I will mark it as pass. Wait, let me double check the 'adversarial' nature. Is there a subtle error in line 1? 'm² + bm + c = 0'. For y'' + p y' + q y = 0, the char eq is m² + pm + q = 0. Here p = -1, q = -2. So m² - m - 2 = 0. The sentence uses 'b' and 'c'. This is standard notation. I don't see an error. I will output pass.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The characteristic equation is incorrectly stated as m² + bm + c = 0 without specifying b and c, leading to a sign error in the roots (should be m² - m - 2 = 0, giving roots 2 and -1, but the text implies a generic form that doesn't match the specific coefficients). More critically, the solution provided in the stated answer and checked in line 5 is 2*(-exp(3*x) - 2)*exp(-x)/3, which simplifies to -2/3 * exp(2x) + 4/3 * exp(-x). Let's check the roots: The characteristic equation m^2 - m - 2 = 0 has roots m = 2 and m = -1. The general solution is y = C1 e^(2x) + C2 e^(-x). y(0) = C1 + C2 = -2. y'(0) = 2C1 - C2 = 0 => C2 = 2C1. 3C1 = -2 => C1 = -2/3, C2 = 4/3. So y = -2/3 e^(2x) + 4/3 e^(-x). The stated answer is 2/3 * (-e^(2x) + 2e^(-x)) = -2/3 e^(2x) + 4/3 e^(-x). This matches. However, line 1 says 'the characteristic equation is m² + bm + c = 0'. For y'' - y' - 2y = 0, the equation is m² - m - 2 = 0. Line 2 shows (m-2)(m+1) which is m² - m - 2. The roots are 2 and -1. Line 3 says 'Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x)'. This is correct. The error is subtle: Line 1 is a generic statement, but it doesn't explicitly link b=-1, c=-2. Is this an error? It's a bit loose. Let's look closer. The prompt asks to find what is wrong. The solution steps are: 1. Setup char eq. 2. Factor/Roots. 3. General form. 4. Apply ICs. 5-7. Verification. The verification lines 5, 6, 7 check the *stated answer* against the ODE and ICs. They pass. The derivation in 1-4 is sketched. Is there a logical gap? Line 1 says 'Try y = e^(mx): the characteristic equation is m² + bm + c = 0'. This is a standard method description. It doesn't explicitly state b=-1, c=-2, but line 2 provides the specific polynomial. This is acceptable shorthand. Wait, look at line 2: '(m - 2)*(m + 1), m**2 - m - 2'. This confirms the polynomial. The roots are 2, -1. The general solution is C1 e^(2x) + C2 e^(-x). The constants are found to be -2/3 and 4/3. The stated answer is equivalent. The solution seems correct. Why would it fail? Let's re-read carefully. 'm² + bm + c = 0'. If the student writes this, they must identify b and c. The problem is y'' - 1y' - 2y = 0. So b = -1, c = -2. The sentence in line 1 is generic. It is not wrong, just generic. Is there a sign error in the roots? (m-2)(m+1) = m^2 - m - 2. Roots 2, -1. Correct. General solution form correct. Constants calculation is skipped but verified by CAS. The verification lines 5, 6, 7 confirm the final answer is correct. Is there a 'misleading' aspect? Line 1 might be considered 'style' if it's too vague, but it's not 'error'. However, often in these adversarial reviews, if the derivation is skipped and only the final answer is verified, it might be considered 'pass' if the verification is robust. But wait, look at line 5. It checks the ODE. Line 6 checks y(0). Line 7 checks y'(0). All proved. The sentences 1-4 describe the method. Is there a flaw in the method description? No. Is there a flaw in the setup? No. Maybe the 'error' is that the solution doesn't actually *show* the calculation of C1 and C2, just says 'Use y(0)... to find C1 and C2' and then jumps to verification. This is a 'style' issue or 'misleading' if it implies the verification is the derivation. But the prompt says 'Review worked calculus solutions'. If the work is missing, is it a fail? Usually, 'pass' is for sound solutions. A solution that skips the algebraic step of solving for constants but verifies the result is technically 'correct' but poor style. However, looking at similar examples, if the logic is sound and the result is verified, it often passes. Let's look for a harder error. Did I miss a sign? y'' - y' - 2y = 0. Char eq: r^2 - r - 2 = 0. (r-2)(r+1)=0. r=2, -1. y = c1 e^(2x) + c2 e^(-x). y(0) = c1+c2 = -2. y'(0) = 2c1 - c2 = 0. c2 = 2c1. 3c1 = -2 -> c1 = -2/3. c2 = 4/3. y = -2/3 e^(2x) + 4/3 e^(-x). Stated answer: 2*(-exp(3*x) - 2)*exp(-x)/3 = 2/3 * (-e^(2x) + 2e^(-x)) = -2/3 e^(2x) + 4/3 e^(-x). It matches. The solution is correct. Why would it be a fail? Perhaps the sentence 'Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x)' is considered incomplete because it doesn't specify that r1 and r2 are the roots found? No, that's implied. Is there a typo in the problem statement vs solution? No. I will mark it as pass. Wait, let me double check the 'adversarial' nature. Is there a subtle error in line 1? 'm² + bm + c = 0'. For y'' + p y' + q y = 0, the char eq is m² + pm + q = 0. Here p = -1, q = -2. So m² - m - 2 = 0. The sentence uses 'b' and 'c'. This is standard notation. I don't see an error. I will output pass.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution skips the step of solving for the constants C1 and C2 using the initial conditions. It jumps from the general form to the specific answer without showing the system of equations or the solution process, making it impossible to verify the logic or learn the method.
gpt-oss:20b: pass 2026-09-28

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/second_order_linear, checked 2026-09-28 with SymPy 1.14.0.