Second-order linear equations
Problem 6.87 · medium
Solve \( \displaystyle y'' - 4y' + 3y = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = 2 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 3\right) \left(m - 1\right) = m^{2} - 4 m + 3 \]Roots: 3, 1.✓ Proved
- Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ 6 e^{x} - 4 \frac{d}{d x} 2 e^{x} + \frac{d^{2}}{d x^{2}} 2 e^{x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 2 \]y(0).
- \[ 2 \]y'(0).
Answer \( y = 2 e^{x} \)
Lines: 2 proved, 5 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
| 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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 7 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 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: fail (misleading) — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions; it merely plugs the final guess 2e^x into the differential equation, which could mislead a student into thinking the constants were derived correctly.qwen3.6:27b-mlx: fail (error) — The solution claims the answer is 2*exp(x) without showing the derivation of the constants C1 and C2 from the initial conditions. The correct solution is y = e^x + e^(3x), not 2*exp(x). The verification steps (5-7) only check that 2*exp(x) satisfies the ODE and one initial condition, but fail to check the second initial condition y'(0)=2 (since (2e^x)' at 0 is 2, wait, 2e^0=2, derivative is 2e^0=2. Actually 2e^x satisfies y(0)=2 and y'(0)=2. Let's re-check the ODE. y=2e^x, y'=2e^x, y''=2e^x. LHS: 2e^x - 4(2e^x) + 3(2e^x) = 2e^x - 8e^x + 6e^x = 0. So 2e^x IS a solution to the ODE and satisfies the ICs? No. The general solution is C1 e^x + C2 e^(3x). y(0) = C1 + C2 = 2. y'(0) = C1 + 3C2 = 2. Subtracting: 2C2 = 0 => C2=0, C1=2. So y=2e^x IS the correct solution. Why did I think it was wrong? Ah, I misremembered the standard example. Let's re-verify. Roots 1, 3. y = c1 e^x + c2 e^(3x). y(0)=c1+c2=2. y'(0)=c1+3c2=2. c1+c2=2, c1+3c2=2 => 2c2=0 => c2=0, c1=2. So y=2e^x is correct. The solution is actually correct. But the text says 'Stated answer: 2*exp(x)' and the solution steps 5-7 verify it. Step 1-3 are generic. Step 4 says 'Use y(0) and y'(0) to find C1 and C2' but doesn't show the work. Is this an error? The prompt asks to review the solution. The solution provided is a sequence of sentences and equations. The equations 5, 6, 7 verify the final answer. The sentences 1-4 describe the method. The method is correct. The answer is correct. The verification is correct. Is there a missing step? Step 4 says 'Use ... to find ...' but doesn't show the finding. However, the final answer is stated and verified. Usually, 'solve' requires showing the work. But the prompt says 'Each line ... is either an EQUATION ... or a SENTENCE'. The solution provided is a mix. It skips the algebra for C1, C2. Is that an error? It's a gap in the derivation. But the final answer is correct and verified. Let's look closer at the prompt's instruction: 'Check that ... each conclusion follows from the lines before it'. Does 'y = 2*exp(x)' follow from lines 1-4? No, lines 1-4 just set up the method. The actual calculation is missing. The verification (5-7) confirms the answer is correct, but the derivation is skipped. Is skipping the derivation an 'error' or 'misleading'? It's incomplete. But is it 'wrong'? The stated answer is correct. The verification proves it is correct. The missing step is just the algebra. I will mark it as 'pass' because the final result is correct and verified, and the method described is correct. Wait, looking at line 1: 'm^2 + bm + c = 0'. The problem is y'' - 4y' + 3y = 0. So b=-4, c=3. Line 2 shows (m-3)(m-1) = m^2 - 4m + 3. This is consistent. Line 3 is correct. Line 4 is a statement of intent. Lines 5-7 verify the specific answer. The solution is technically 'sound' in that the answer is correct and the method is valid, even if the intermediate algebra is omitted. However, usually, a solution should show the work. But the constraints say 'do not re-check the algebra of a line marked proved'. The lines 5-7 are marked proved. They verify the answer. The sentences 1-4 are correct. I will pass it.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution claims the answer is 2*exp(x) without showing the derivation of the constants C1 and C2 from the initial conditions. The correct solution is y = e^x + e^(3x), not 2*exp(x). The verification steps (5-7) only check that 2*exp(x) satisfies the ODE and one initial condition, but fail to check the second initial condition y'(0)=2 (since (2e^x)' at 0 is 2, wait, 2e^0=2, derivative is 2e^0=2. Actually 2e^x satisfies y(0)=2 and y'(0)=2. Let's re-check the ODE. y=2e^x, y'=2e^x, y''=2e^x. LHS: 2e^x - 4(2e^x) + 3(2e^x) = 2e^x - 8e^x + 6e^x = 0. So 2e^x IS a solution to the ODE and satisfies the ICs? No. The general solution is C1 e^x + C2 e^(3x). y(0) = C1 + C2 = 2. y'(0) = C1 + 3C2 = 2. Subtracting: 2C2 = 0 => C2=0, C1=2. So y=2e^x IS the correct solution. Why did I think it was wrong? Ah, I misremembered the standard example. Let's re-verify. Roots 1, 3. y = c1 e^x + c2 e^(3x). y(0)=c1+c2=2. y'(0)=c1+3c2=2. c1+c2=2, c1+3c2=2 => 2c2=0 => c2=0, c1=2. So y=2e^x is correct. The solution is actually correct. But the text says 'Stated answer: 2*exp(x)' and the solution steps 5-7 verify it. Step 1-3 are generic. Step 4 says 'Use y(0) and y'(0) to find C1 and C2' but doesn't show the work. Is this an error? The prompt asks to review the solution. The solution provided is a sequence of sentences and equations. The equations 5, 6, 7 verify the final answer. The sentences 1-4 describe the method. The method is correct. The answer is correct. The verification is correct. Is there a missing step? Step 4 says 'Use ... to find ...' but doesn't show the finding. However, the final answer is stated and verified. Usually, 'solve' requires showing the work. But the prompt says 'Each line ... is either an EQUATION ... or a SENTENCE'. The solution provided is a mix. It skips the algebra for C1, C2. Is that an error? It's a gap in the derivation. But the final answer is correct and verified. Let's look closer at the prompt's instruction: 'Check that ... each conclusion follows from the lines before it'. Does 'y = 2*exp(x)' follow from lines 1-4? No, lines 1-4 just set up the method. The actual calculation is missing. The verification (5-7) confirms the answer is correct, but the derivation is skipped. Is skipping the derivation an 'error' or 'misleading'? It's incomplete. But is it 'wrong'? The stated answer is correct. The verification proves it is correct. The missing step is just the algebra. I will mark it as 'pass' because the final result is correct and verified, and the method described is correct. Wait, looking at line 1: 'm^2 + bm + c = 0'. The problem is y'' - 4y' + 3y = 0. So b=-4, c=3. Line 2 shows (m-3)(m-1) = m^2 - 4m + 3. This is consistent. Line 3 is correct. Line 4 is a statement of intent. Lines 5-7 verify the specific answer. The solution is technically 'sound' in that the answer is correct and the method is valid, even if the intermediate algebra is omitted. However, usually, a solution should show the work. But the constraints say 'do not re-check the algebra of a line marked proved'. The lines 5-7 are marked proved. They verify the answer. The sentences 1-4 are correct. I will pass it.gpt-oss:20b: fail (misleading) 2026-09-28 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions; it merely plugs the final guess 2e^x into the differential equation, which could mislead a student into thinking the constants were derived correctly.qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution claims the answer is 2*exp(x) without showing the derivation of the constants C1 and C2 from the initial conditions. While the final answer is numerically correct, the logical step of solving for the constants is missing, making the solution incomplete and potentially misleading about the method.gpt-oss:20b: fail (misleading) 2026-09-27 — The solution jumps to y=2e^x without solving for the constants C₁ and C₂ from the initial conditions; it merely checks that this guess satisfies the differential equation and the ICs, which is not a valid derivation of the unique solution.qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions, instead jumping to a specific answer (2*exp(x)) which is incorrect (the correct solution is e^x + e^(3x)). The verification steps only check that the proposed answer satisfies the initial conditions, but do not show the derivation.gpt-oss:20b: pass 2026-09-26
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 2026-10-13 with SymPy 1.14.0.