∫Calc Practice

Undetermined coefficients

Problem 6.369 · hard

Solve \( \displaystyle y'' + 3y' + 2y = -1 \) with \( \displaystyle y(0) = -2 \), \( \displaystyle y'(0) = 1 \).
  1. \[ r^{2} + 3 r + 2 = \left(r + 1\right) \left(r + 2\right) \]
    The characteristic equation has roots -1 and -2.✓ Proved
  2. So y_h = C₁e^(-1x) + C₂e^(-2x). Guess y_p = A.
  3. \[ 3 \frac{d}{d x} \left(- \frac{1}{2}\right) + \frac{d^{2}}{d x^{2}} \left(- \frac{1}{2}\right) - 1 = -1 \]
    Matching coefficients gives y_p = -1/2; it satisfies the equation.✓ Proved
  4. \[ \left[\begin{matrix}\left. - \frac{1}{2} - 2 e^{- x} + \frac{e^{- 2 x}}{2} \right|_{\substack{ x=0 }}\\\left. \frac{d}{d x} \left(- \frac{1}{2} - 2 e^{- x} + \frac{e^{- 2 x}}{2}\right) \right|_{\substack{ x=0 }}\end{matrix}\right] = \left[\begin{matrix}-2\\1\end{matrix}\right] \]
    The initial conditions fix C₁ = -2 and C₂ = 1/2.✓ Proved
Answer \( y = - \frac{1}{2} - 2 e^{- x} + \frac{e^{- 2 x}}{2} \)

Lines: 3 proved, 1 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0SymPy's dsolve with the same initial conditions agrees

Reviewers

  • gpt-oss:20b: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The algebraic check for the particular solution is incorrect: for a constant \(A\), \(y''+3y'+2y=2A\), so \(2A=-1\) giving \(A=-\tfrac12\). The provided e
  • qwen3.6:27b-mlx: pass — The solution correctly identifies the homogeneous and particular solutions, applies the initial conditions accurately, and arrives at the correct final answer.
Every verdict on record (4)
  • gpt-oss:20b: inconclusive 2026-10-11 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The algebraic check for the particular solution is incorrect: for a constant \(A\), \(y''+3y'+2y=2A\), so \(2A=-1\) giving \(A=-\tfrac12\). The provided e
  • qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly identifies the homogeneous and particular solutions, applies the initial conditions accurately, and arrives at the correct final answer.
  • qwen3.6:27b-mlx: fail (misleading) 2026-10-11 — The solution jumps from the general form to the final answer without showing the derivation of the constants C1 and C2 from the initial conditions. While the final answer is correct, the sentence in step 4 claims the matrix equality fixes the constants, but it does not explicitly show the system of equations or the solution process, which is a critical step for a student to learn. Furthermore, step 3's notation is confusing and non-standard.
  • gpt-oss:20b: pass 2026-10-11

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/undetermined_coefficients, checked 2026-10-11 with SymPy 1.14.0.