Second-order linear equations
Problem 6.110 · medium
Solve \( \displaystyle y'' - 2y' + 1y = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = -3 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 1\right)^{2} = m^{2} - 2 m + 1 \]Roots: 1.✓ Proved
- A repeated root r gives y = (C₁ + C₂x)e^(rx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ - 3 x e^{x} - 2 \frac{d}{d x} \left(- 3 x e^{x}\right) + \frac{d^{2}}{d x^{2}} \left(- 3 x e^{x}\right) = 0 \]The solution satisfies the equation.✓ Proved
- \[ 0 \]y(0).✓ Proved
- \[ -3 \]y'(0).✓ Proved
Answer \( y = - 3 x e^{x} \)
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: passqwen3.6:27b-mlx: fail (error) — The characteristic equation is m^2 - 2m + 1 = 0, which has a repeated root m=1. The general solution is y = (C1 + C2x)e^x. Applying y(0)=0 gives C1=0. Then y' = C2(e^x + xe^x), so y'(0) = C2 = -3. The correct solution is y = -3xe^x. However, the stated answer in the problem description is -3*x*exp(x), which matches the derived solution. Wait, let me re-read the problem statement carefully. Problem: Solve y'' - 2y' + 1y = 0. Characteristic eq: r^2 - 2r + 1 = 0 => (r-1)^2 = 0 => r=1. General solution: y = c1 e^x + c2 x e^x. y(0) = c1 = 0. y' = c2(e^x + x e^x). y'(0) = c2 = -3. So y = -3 x e^x. The stated answer is -3*x*exp(x). The solution steps are correct. Why did I think it was wrong? Let me check the sign. y'' - 2y' + y = 0. If y = -3xe^x, y' = -3(e^x + xe^x) = -3e^x(1+x). y'' = -3(e^x(1+x) + e^x) = -3e^x(2+x). y'' - 2y' + y = -3e^x(2+x) - 2(-3e^x(1+x)) - 3xe^x = -3e^x [ 2+x - 2(1+x) + x ] = -3e^x [ 2+x - 2 - 2x + x ] = -3e^x [ 0 ] = 0. It works. The solution is correct. The verdict should be pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The characteristic equation is m^2 - 2m + 1 = 0, which has a repeated root m=1. The general solution is y = (C1 + C2x)e^x. Applying y(0)=0 gives C1=0. Then y' = C2(e^x + xe^x), so y'(0) = C2 = -3. The correct solution is y = -3xe^x. However, the stated answer in the problem description is -3*x*exp(x), which matches the derived solution. Wait, let me re-read the problem statement carefully. Problem: Solve y'' - 2y' + 1y = 0. Characteristic eq: r^2 - 2r + 1 = 0 => (r-1)^2 = 0 => r=1. General solution: y = c1 e^x + c2 x e^x. y(0) = c1 = 0. y' = c2(e^x + x e^x). y'(0) = c2 = -3. So y = -3 x e^x. The stated answer is -3*x*exp(x). The solution steps are correct. Why did I think it was wrong? Let me check the sign. y'' - 2y' + y = 0. If y = -3xe^x, y' = -3(e^x + xe^x) = -3e^x(1+x). y'' = -3(e^x(1+x) + e^x) = -3e^x(2+x). y'' - 2y' + y = -3e^x(2+x) - 2(-3e^x(1+x)) - 3xe^x = -3e^x [ 2+x - 2(1+x) + x ] = -3e^x [ 2+x - 2 - 2x + x ] = -3e^x [ 0 ] = 0. It works. The solution is correct. The verdict should be pass.gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: fail (error) 2026-09-26 — The characteristic equation for y'' - 2y' + y = 0 is m^2 - 2m + 1 = 0, which has a repeated root m = 1. The general solution is y = (C1 + C2x)e^x. Applying y(0) = 0 gives C1 = 0. Then y' = C2(e^x + xe^x), so y'(0) = C2 = -3. The correct solution is y = -3xe^x. However, the stated answer in the problem description is -3*x*exp(x), which matches. Wait, let me re-read the prompt's stated answer. The prompt says 'Stated answer: -3*x*exp(x)'. The solution derives this. But look at line 1: 'm^2 + bm + c = 0'. For y'' - 2y' + y = 0, b = -2, c = 1. The equation is m^2 - 2m + 1 = 0. Line 2 shows '(m - 1)**2, m**2 - 2*m + 1'. This is correct. Line 3 says 'A repeated root r gives y = (C1 + C2x)e^(rx)'. This is correct. Line 4 says 'Use y(0) and y'(0) to find C1 and C2'. This is correct. Lines 5-7 verify the result. The solution seems correct. Why would it fail? Let me check the signs. y = -3xe^x. y' = -3(e^x + xe^x) = -3e^x(1+x). y'(0) = -3(1) = -3. Correct. y(0) = 0. Correct. y'' = -3(e^x(1+x) + e^x) = -3e^x(2+x). y'' - 2y' + y = -3e^x(2+x) - 2(-3e^x(1+x)) + (-3xe^x) = -3e^x [ 2+x - 2(-1-x) - x ] = -3e^x [ 2+x + 2 + 2x - x ] = -3e^x [ 4 + 2x ]? No. Let's recompute. y'' - 2y' + y. y = -3xe^x. y' = -3e^x - 3xe^x. y'' = -3e^x - 3e^x - 3xe^x = -6e^x - 3xe^x. y'' - 2y' + y = (-6e^x - 3xe^x) - 2(-3e^x - 3xe^x) + (-3xe^x) = -6e^x - 3xe^x + 6e^x + 6xe^x - 3xe^x = 0. It works. The solution is correct. Is there a subtle error? Line 1 says 'm^2 + bm + c = 0'. It doesn't specify b and c. Line 2 shows the specific polynomial. This is fine. The verdict should be pass.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 2026-09-26 with SymPy 1.14.0.