First-order linear equations
Problem 6.68 · hard
Solve \( \displaystyle y' + 3y = 1 \) with \( \displaystyle y(0) = 1 \).
- The equation is linear in standard form; the integrating factor is e^(∫3 dx) = e^(3x).
- \[ \frac{d}{d x} Y{\left(x \right)} e^{3 x} = 3 Y{\left(x \right)} e^{3 x} + e^{3 x} \frac{d}{d x} Y{\left(x \right)} \]Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
- \[ \int e^{3 x}\, dx = \frac{e^{3 x}}{3} \]Integrate the right side.✓ Proved
- Setting x = 0 and y = 1 fixes the constant of integration: C = 2/3.
- \[ \frac{d}{d x} \left(\frac{1}{3} + \frac{2 e^{- 3 x}}{3}\right) + 1 + 2 e^{- 3 x} = 1 \]The solution satisfies the equation.✓ Proved
- \[ 1 \]And the initial condition.✓ Proved
Answer \( y = \frac{1}{3} + \frac{2 e^{- 3 x}}{3} \)
Lines: 4 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 | ✓ 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; y(0) matches |
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 2026-09-26 with SymPy 1.14.0.