First-order linear equations
Problem 6.24 · medium
Solve \( \displaystyle y' + 3y = e^{- 3 x} \) with \( \displaystyle y(0) = 3 \).
- 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
- \[ \int 1\, dx = x \]Integrate the right side.✓ Proved
- Setting x = 0 and y = 3 fixes the constant of integration: C = 3.
- \[ \left(3 x + 9\right) e^{- 3 x} + \frac{d}{d x} \left(x + 3\right) e^{- 3 x} = e^{- 3 x} \]The solution satisfies the equation.✓ Proved
- \[ 3 \]And the initial condition.✓ Proved
Answer \( y = \left(x + 3\right) e^{- 3 x} \)
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.