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Second-order initial and boundary value problems

Problem 6.805 · medium

Solve \( \displaystyle - 2 y - y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 3 \).
  1. \[ m^{2} - m - 2 = \left(m - 2\right) \left(m + 1\right) \]
    The characteristic equation has distinct real roots 2 and -1.✓ Proved
  2. So y = C₁exp(2*x) + C₂exp(-x); the two conditions fix C₁ and C₂.
  3. \[ \frac{\left(14 - 2 e^{3 x}\right) e^{- x}}{3} - \frac{d}{d x} \frac{\left(e^{3 x} - 7\right) e^{- x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(e^{3 x} - 7\right) e^{- x}}{3} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{\left(e^{3 x} - 7\right) e^{- x}}{3} \right|_{\substack{ x=0 }} = -2 \]
    ... y(0) = -2 ...✓ Proved
  5. \[ \left. \frac{d}{d x} \frac{\left(e^{3 x} - 7\right) e^{- x}}{3} \right|_{\substack{ x=0 }} = 3 \]
    y′(0) = 3.✓ Proved
Answer \( y = \frac{\left(e^{3 x} - 7\right) e^{- x}}{3} \)

Lines: 4 proved, 1 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0finite differences of the answer satisfy the equation, and the conditions hold in floating point

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The characteristic equation derived in step 1 (m^2 - m - 2 = 0) does not match the differential equation y'' - y' - 2y = 0, which yields m^2 - m - 2 = 0. Wait, let's re-read carefully. The ODE is y'' - y' - 2y = 0. The characteristic equation is r^2 - r - 2 = 0. The roots are r = 2 and r = -1. Step 1 says m^2 - m - 2 = (m-2)(m+1). This is correct algebra. Step 2 says y = C1 e^(2x) + C2 e^(-x). This is the correct general solution. However, the stated answer is y = (e^(3x) - 7)e^(-x)/3 = (1/3)e^(2x) - (7/3)e^(-x). Let's check if this satisfies the initial conditions. y(0) = 1/3 - 7/3 = -6/3 = -2. Correct. y'(x) = (2/3)e^(2x) + (7/3)e^(-x). y'(0) = 2/3 + 7/3 = 9/3 = 3. Correct. So the stated answer is correct. The solution steps 3, 4, 5 verify the stated answer. Step 1 and 2 set up the general method. Is there an error? Step 1 says 'The characteristic equation has distinct real roots 2 and -1'. This is true. Step 2 says 'So y = ...'. This is true. The solution is logically sound and the answer is correct. Why would it fail? Let's look closer at the prompt's instruction: 'find what is wrong'. Is there a subtle error? The problem statement is '- 2 y - y' + y'' = 0'. This is y'' - y' - 2y = 0. Characteristic eq: r^2 - r - 2 = 0. Roots: 2, -1. General solution: y = c1 e^(2x) + c2 e^(-x). The provided answer is a specific instance of this. The verification steps are correct. The setup is correct. It seems to pass. Wait, look at step 1 again. 'm**2 - m - 2 = (m - 2)*(m + 1)'. This is an identity. It doesn't explicitly state 'The characteristic equation is...'. It just shows a factorization. But the text says 'The characteristic equation has distinct real roots 2 and -1'. This implies the characteristic equation is m^2 - m - 2 = 0. This is correct. Is there a sign error in the problem statement interpretation? '- 2 y - y' + y'' = 0' -> y'' - y' - 2y = 0. Correct. What if the problem was y'' + y' - 2y = 0? Then roots would be 1, -2. But the text matches the equation. Let's look at the answer format. y = (e^(3x) - 7)e^(-x)/3. This simplifies to e^(2x)/3 - 7e^(-x)/3. This is of the form c1 e^(2x) + c2 e^(-x). It is correct. I see no error. Pass.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-11-07 — The characteristic equation derived in step 1 (m^2 - m - 2 = 0) does not match the differential equation y'' - y' - 2y = 0, which yields m^2 - m - 2 = 0. Wait, let's re-read carefully. The ODE is y'' - y' - 2y = 0. The characteristic equation is r^2 - r - 2 = 0. The roots are r = 2 and r = -1. Step 1 says m^2 - m - 2 = (m-2)(m+1). This is correct algebra. Step 2 says y = C1 e^(2x) + C2 e^(-x). This is the correct general solution. However, the stated answer is y = (e^(3x) - 7)e^(-x)/3 = (1/3)e^(2x) - (7/3)e^(-x). Let's check if this satisfies the initial conditions. y(0) = 1/3 - 7/3 = -6/3 = -2. Correct. y'(x) = (2/3)e^(2x) + (7/3)e^(-x). y'(0) = 2/3 + 7/3 = 9/3 = 3. Correct. So the stated answer is correct. The solution steps 3, 4, 5 verify the stated answer. Step 1 and 2 set up the general method. Is there an error? Step 1 says 'The characteristic equation has distinct real roots 2 and -1'. This is true. Step 2 says 'So y = ...'. This is true. The solution is logically sound and the answer is correct. Why would it fail? Let's look closer at the prompt's instruction: 'find what is wrong'. Is there a subtle error? The problem statement is '- 2 y - y' + y'' = 0'. This is y'' - y' - 2y = 0. Characteristic eq: r^2 - r - 2 = 0. Roots: 2, -1. General solution: y = c1 e^(2x) + c2 e^(-x). The provided answer is a specific instance of this. The verification steps are correct. The setup is correct. It seems to pass. Wait, look at step 1 again. 'm**2 - m - 2 = (m - 2)*(m + 1)'. This is an identity. It doesn't explicitly state 'The characteristic equation is...'. It just shows a factorization. But the text says 'The characteristic equation has distinct real roots 2 and -1'. This implies the characteristic equation is m^2 - m - 2 = 0. This is correct. Is there a sign error in the problem statement interpretation? '- 2 y - y' + y'' = 0' -> y'' - y' - 2y = 0. Correct. What if the problem was y'' + y' - 2y = 0? Then roots would be 1, -2. But the text matches the equation. Let's look at the answer format. y = (e^(3x) - 7)e^(-x)/3. This simplifies to e^(2x)/3 - 7e^(-x)/3. This is of the form c1 e^(2x) + c2 e^(-x). It is correct. I see no error. Pass.
  • gpt-oss:20b: pass 2026-11-07
  • qwen3.6:27b-mlx: fail (error) 2026-11-06 — The characteristic equation derived in step 1 (m^2 - m - 2) corresponds to y'' - y' - 2y = 0, but the problem states -2y - y' + y'' = 0, which is equivalent to y'' - y' - 2y = 0. Wait, let me re-read carefully. The problem is -2y - y' + y'' = 0. Rearranging terms: y'' - y' - 2y = 0. The characteristic equation is indeed m^2 - m - 2 = 0. The roots are 2 and -1. The general solution is y = C1 e^(2x) + C2 e^(-x). The stated answer is y = (e^(3x) - 7)e^(-x)/3 = (1/3)e^(2x) - (7/3)e^(-x). Let's check the initial conditions with this answer. y(0) = (1/3) - (7/3) = -6/3 = -2. Correct. y'(x) = (2/3)e^(2x) + (7/3)e^(-x). y'(0) = 2/3 + 7/3 = 9/3 = 3. Correct. The solution steps verify the final answer directly rather than solving for C1 and C2, but the verification is correct. However, step 1 says 'The characteristic equation has distinct real roots 2 and -1'. Step 2 says 'So y = C1 exp(2x) + C2 exp(-x)'. This is correct. Steps 3-5 verify the specific answer. The logic is sound, just skipping the algebra of finding C1 and C2. Is there an error? Let's look at the problem statement again. '- 2 y - y' + y'' = 0'. This is y'' - y' - 2y = 0. Characteristic eq: r^2 - r - 2 = 0. (r-2)(r+1)=0. Roots 2, -1. General solution y = c1 e^{2x} + c2 e^{-x}. The provided answer is y = 1/3 e^{2x} - 7/3 e^{-x}. This matches the form. The verification steps are correct. Why would it fail? Perhaps the 'sentence' in step 2 implies a derivation that isn't shown, but it's not false. It says 'the two conditions fix C1 and C2'. Then it jumps to verifying the answer. This is a valid way to present a solution if the answer is given or derived elsewhere. It's not 'wrong'. Is it misleading? It skips the actual calculation of C1 and C2. But the prompt asks to review 'worked calculus solutions'. If the solution doesn't show the work of finding the constants, is it an error? The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Step 2 is a sentence. It claims the conditions fix the constants. Steps 3-5 verify the result. This is logically sound. I will mark it as pass.
  • gpt-oss:20b: pass 2026-11-06
  • qwen3.6:27b-mlx: fail (error) 2026-11-06 — The characteristic equation is derived incorrectly; for y'' - y' - 2y = 0, the equation is m^2 - m - 2 = 0, which factors as (m-2)(m+1)=0, giving roots 2 and -1. However, the stated answer y = (e^{3x} - 7)e^{-x}/3 simplifies to (1/3)e^{2x} - (7/3)e^{-x}, which corresponds to roots 2 and -1. Wait, let's re-read the problem statement carefully. The problem is -2y - y' + y'' = 0, which is y'' - y' - 2y = 0. The roots are indeed 2 and -1. The general solution is C1 e^{2x} + C2 e^{-x}. The stated answer is (1/3)e^{2x} - (7/3)e^{-x}. Let's check the initial conditions. y(0) = 1/3 - 7/3 = -6/3 = -2. Correct. y'(x) = (2/3)e^{2x} + (7/3)e^{-x}. y'(0) = 2/3 + 7/3 = 9/3 = 3. Correct. So the answer is correct. Why did I think it was wrong? Let me re-read the solution steps. Step 1 says m^2 - m - 2 = (m-2)(m+1). This is correct. Step 2 says y = C1 exp(2x) + C2 exp(-x). This is correct. Steps 3-5 verify the answer. The solution is actually correct. Why would it fail? Let me look closer at the prompt's instruction. "Review worked calculus solutions adversarially." Is there a subtle error? The problem is -2y - y' + y'' = 0. Standard form y'' - y' - 2y = 0. Characteristic eq: r^2 - r - 2 = 0. Roots: (1 +/- sqrt(1+8))/2 = (1 +/- 3)/2. r1 = 2, r2 = -1. General solution: y = c1 e^{2x} + c2 e^{-x}. y(0) = c1 + c2 = -2. y'(0) = 2c1 - c2 = 3. Adding equations: 3c1 = 1 => c1 = 1/3. c2 = -2 - 1/3 = -7/3. Solution: y = (1/3)e^{2x} - (7/3)e^{-x} = (e^{2x} - 7e^{-x})/3? No, the stated answer is (e^{3x} - 7)e^{-x}/3 = (e^{2x} - 7e^{-x})/3. Yes, e^{3x} * e^{-x} = e^{2x}. So the answer is correct. The steps are correct. The verification steps are correct. Is there a style issue? Step 1 is marked [equation, unchecked] but contains text "The characteristic equation has distinct real roots 2 and -1." This is a sentence mixed with an equation. But the prompt says "Each line... is either an EQUATION... or a SENTENCE". Line 1 is labeled [equation, unchecked]. It contains an equation and a comment. The comment is correct. Line 2 is a sentence. It is correct. Lines 3-5 are equations verifying the answer. They are correct. So the solution is sound. Verdict: pass.
  • gpt-oss:20b: pass 2026-11-06

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_ivp_bvp, checked 2026-11-07 with SymPy 1.14.0.

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