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First-order linear equations

Problem 6.105 · medium Video

Solve \( \displaystyle y' + 1y = e^{- x} \) with \( \displaystyle y(0) = -3 \).

The equation is linear in standard form; the integrating factor is e^(∫1 dx) = e^(1x).
\[ \frac{d}{d x} Y{\left(x \right)} e^{x} = Y{\left(x \right)} e^{x} + e^{x} \frac{d}{d x} Y{\left(x \right)} \]
Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
\[ \int 1\, dx = x \]
Integrate the right side.✓ Proved
Setting x = 0 and y = -3 fixes the constant of integration: C = -3.
\[ \left(x - 3\right) e^{- x} + \frac{d}{d x} \left(x - 3\right) e^{- x} = e^{- x} \]
The solution satisfies the equation.✓ Proved
\[ -3 \]
And the initial condition.

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

The slope field of y' = -y + e^(-x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -3) is highlighted, y = (x - 3)·e^(-x).

Lines: 3 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. 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
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
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	sympy.checkodesol substitutes the solution back; y(0) matches

Reviewers

gpt-oss:20b: fail (error) — The constant of integration is mis‑applied: with C = -3 the solution should be y = (-3 - x)e^{-x} = -(x+3)e^{-x}, not (x-3)e^{-x}. The final answer is incorrect.
qwen3.6:27b-mlx: pass — The solution correctly identifies the integrating factor, performs the integration, applies the initial condition, and verifies the result. The steps are logically sound and mathematically correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the integrating factor, performs the integration, applies the initial condition, and verifies the result. The steps are logically sound and mathematically correct.
gpt-oss:20b: fail (error) 2026-09-27 — The constant of integration is mis‑applied: with C = -3 the solution should be y = (-3 - x)e^{-x} = -(x+3)e^{-x}, not (x-3)e^{-x}. The final answer is incorrect.
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly derive the general solution Y(x) = (x + C)e^{-x} from the integrated equation. It jumps from the integral of the RHS to the specific constant C without showing the intermediate step of dividing by the integrating factor, making the logical flow incomplete and potentially confusing for a student.
gpt-oss:20b: pass 2026-09-26
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the integrating factor, applies the method of integrating factors, and correctly determines the constant of integration to satisfy the initial condition.
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/linear_first_order, checked 2026-10-13 with SymPy 1.14.0.