Home›Calculus 2›First-order linear equations›Problem 6.22

First-order linear equations

Problem 6.22 · hard Video

Solve \( \displaystyle y' + 2y = 2 e^{- 2 x} \) with \( \displaystyle y(0) = 4 \).

The equation is linear in standard form; the integrating factor is e^(∫2 dx) = e^(2x).
\[ \frac{d}{d x} Y{\left(x \right)} e^{2 x} = 2 Y{\left(x \right)} e^{2 x} + e^{2 x} \frac{d}{d x} Y{\left(x \right)} \]
Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
\[ \int 2\, dx = 2 x \]
Integrate the right side.✓ Proved
Setting x = 0 and y = 4 fixes the constant of integration: C = 4.
\[ \left(4 x + 8\right) e^{- 2 x} + \frac{d}{d x} \left(2 x + 4\right) e^{- 2 x} = 2 e^{- 2 x} \]
The solution satisfies the equation.✓ Proved
\[ 4 \]
And the initial condition.

Answer \( y = 2 \left(x + 2\right) e^{- 2 x} \)

The slope field of y' = -2y + 2e^(-2x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 4) is highlighted, y = (2x + 4)·e^(-2x).

Lines: 3 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 proposed solution y=4(x+2)e^{-2x} does not satisfy the differential equation; the correct solution is y=2(x+2)e^{-2x}.
qwen3.6:27b-mlx: fail (error) — The constant of integration is calculated incorrectly. Integrating the right side 2 gives 2x + C. Using y(0)=4 implies 4 = 0 + C, so C=4. The general solution is y = (2x + 4)e^{-2x}. The stated answer is 2(x+2)e^{-2x} = (2x+4)e^{-2x}, which is correct, but line 4 claims C=4 fixes the constant, implying the solution is y = (2x + 4)e^{-2x}. However, line 5 checks 4*(x+2)... wait, line 5 checks 4*(x+2)exp(-2x) which is 4x+8, not 2x+4. The stated answer is 2(x+2)exp(-2x). Line 5 verifies 4(x+2)exp(-2x) which is WRONG. The solution derived from C=4 is y = (2x+4)e^{-2x} = 2(x+2)e^{-2x}. Line 5 checks 4(x+2)e^{-2x} which is double the correct answer. Thus the verification in line 5 is for the wrong function, and the stated answer in the problem description matches the correct derivation, but the solution text's verification step is inconsistent with the stated answer.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The constant of integration is calculated incorrectly. Integrating the right side 2 gives 2x + C. Using y(0)=4 implies 4 = 0 + C, so C=4. The general solution is y = (2x + 4)e^{-2x}. The stated answer is 2(x+2)e^{-2x} = (2x+4)e^{-2x}, which is correct, but line 4 claims C=4 fixes the constant, implying the solution is y = (2x + 4)e^{-2x}. However, line 5 checks 4*(x+2)... wait, line 5 checks 4*(x+2)exp(-2x) which is 4x+8, not 2x+4. The stated answer is 2(x+2)exp(-2x). Line 5 verifies 4(x+2)exp(-2x) which is WRONG. The solution derived from C=4 is y = (2x+4)e^{-2x} = 2(x+2)e^{-2x}. Line 5 checks 4(x+2)e^{-2x} which is double the correct answer. Thus the verification in line 5 is for the wrong function, and the stated answer in the problem description matches the correct derivation, but the solution text's verification step is inconsistent with the stated answer.
gpt-oss:20b: fail (error) 2026-09-28 — The proposed solution y=4(x+2)e^{-2x} does not satisfy the differential equation; the correct solution is y=2(x+2)e^{-2x}.
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The constant of integration is calculated incorrectly. Integrating the right side 2 yields 2x + C. Applying y(0)=4 to y*e^(2x) = 2x + C gives 4 = 0 + C, so C=4. The general solution is y = (2x + 4)e^(-2x) = 2(x+2)e^(-2x). However, line 4 states 'C=4' but the final answer in line 5 uses a coefficient of 2 for the term (x+2), implying the student might be confused about where the factor of 2 comes from or how C relates to the final form. More critically, the solution presented in line 5 is 4*(x+2)*exp(-2*x) in the check, but the stated answer is 2*(x+2)*exp(-2*x). Let's re-evaluate. y' + 2y = 2e^(-2x). IF = e^(2x). (y e^(2x))' = 2. y e^(2x) = 2x + C. y = (2x + C)e^(-2x). y(0) = 4 => C = 4. So y = (2x + 4)e^(-2x) = 2(x+2)e^(-2x). The stated answer is correct. Line 5 checks '4*(x+2)*exp(-2*x)'? No, line 5 says: '4*(x + 2)*exp(-2*x) + Derivative(2*(x + 2)*exp(-2*x), x), 2*exp(-2*x)'. This equation is garbled. It seems to be checking if the derivative of the stated answer plus 2 times the stated answer equals the RHS. But the first term is 4*(x+2)... which is 2*y. And the derivative is of 2*(x+2)... which is y'. So it checks 2y + y' = RHS. But why is the first term 4*(x+2)...? If y = 2(x+2)e^(-2x), then 2y = 4(x+2)e^(-2x). So the equation in line 5 is effectively checking 2y + y' = 2e^(-2x). This is correct algebraically. However, line 4 says 'C=4'. This is correct. The main issue is that line 5 is an opaque verification step that doesn't clearly derive the solution, and line 4 is a bit abrupt. But is it an error? The solution steps are: 1. IF. 2. LHS becomes derivative. 3. Integrate RHS to get 2x. 4. Find C. 5. Verify. 6. Verify IC. The jump from 3 to 4 skips writing the general solution y e^(2x) = 2x + C. It just says 'Setting x=0... fixes C=4'. This is logically sound if one assumes the intermediate step. The stated answer is correct. The verification in 5 is correct. Is there a misleading part? Line 4 says 'C=4'. In the expression y = (2x+C)e^(-2x), C is indeed 4. The final answer is 2(x+2)e^(-2x). This matches. I don't see a hard error. Wait, look at line 5 again. '4*(x + 2)*exp(-2*x) + Derivative(2*(x + 2)*exp(-2*x), x), 2*exp(-2*x)'. This is checking if 2*y + y' = 2e^(-2x). Since y = 2(x+2)e^(-2x), 2y = 4(x+2)e^(-2x). So it is checking the ODE. This is fine. Why would it fail? Maybe the prompt implies the solution *derives* the answer. It doesn't really derive it, it verifies it. But 'Solve' usually requires derivation. However, the instructions say 'check that... each conclusion follows'. Line 4 concludes C=4. Line 5 verifies the answer. It doesn't explicitly state the general solution. Is that an error? It's a gap. But is it 'wrong'? The stated answer is correct. The verification is correct. The constant calculation is correct. I will mark it as pass because the math is correct, even if the derivation is compressed.
gpt-oss:20b: fail (error) 2026-09-27 — The solution mistakenly uses 4*(x+2)e^{-2x} instead of the correct (2x+4)e^{-2x}. This leads to y'+2y=4e^{-2x} rather than the required 2e^{-2x}, so the conclusion is incorrect.
qwen3.6:27b-mlx: inconclusive 2026-09-26 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The constant of integration is calculated incorrectly. The general solution is y = (2x + C)e^(-2x). Applying y(0)=4 gives 4 = (0 + C)*1, so C=4. The
gpt-oss:20b: fail (misleading) 2026-09-26 — The solution incorrectly tests the function 4*(x+2)e^{-2x}, which does not satisfy the differential equation (it yields 4e^{-2x} instead of 2e^{-2x}). The correct solution is 2*(x+2)e^{-2x}. The step setting C=4 is fine, but the subsequent test is misleading.

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.