First-order linear equations
Problem 6.17 · hard
Solve \( \displaystyle y' + 3y = e^{x} \) with \( \displaystyle y(0) = 2 \).
- 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
- \[ \int e^{4 x}\, dx = \frac{e^{4 x}}{4} \]Integrate the right side.✓ Proved
- Setting x = 0 and y = 2 fixes the constant of integration: C = 7/4.
- \[ \frac{\left(3 e^{4 x} + 21\right) e^{- 3 x}}{4} + \frac{d}{d x} \frac{\left(e^{4 x} + 7\right) e^{- 3 x}}{4} = e^{x} \]The solution satisfies the equation.✓ Proved
- \[ 2 \]And the initial condition.✓ Proved
Answer \( y = \frac{\left(e^{4 x} + 7\right) e^{- 3 x}}{4} \)
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.