Home›Calculus 3›Second-order linear equations›Problem 6.42

Second-order linear equations

Problem 6.42 · hard

Solve \( \displaystyle y'' + 2y' - 3y = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 3 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 1\right) \left(m + 3\right) = m^{2} + 2 m - 3 \]
Roots: 1, -3.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \frac{\left(3 - 9 e^{4 x}\right) e^{- 3 x}}{2} + 2 \frac{d}{d x} \frac{\left(3 e^{4 x} - 1\right) e^{- 3 x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(3 e^{4 x} - 1\right) e^{- 3 x}}{2} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 1 \]
y(0).
\[ 3 \]
y'(0).

Answer \( y = \frac{\left(3 e^{4 x} - 1\right) e^{- 3 x}}{2} \)

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: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer, relying on the verification steps (lines 5-7) to prove correctness rather than demonstrating the derivation. A student would not learn how to determine the coefficients from the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer, relying on the verification steps (lines 5-7) to prove correctness rather than demonstrating the derivation. A student would not learn how to determine the coefficients from the initial conditions.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly solve for the constants C1 and C2, jumping directly to the final answer without showing the system of equations derived from the initial conditions. This omits the core logical step of the problem.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the characteristic equation, finds the roots, states the general form, and verifies the specific solution against the differential equation and initial conditions. The steps are logically sound and the final answer is correct.
gpt-oss:20b: fail (error) 2026-09-26 — The solution never actually determines the constants C₁ and C₂ from the initial conditions; it only verifies that a proposed expression satisfies the differential equation and the initial values. The derivation step is missing, so the reasoning is incomplete.

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.