Second-order linear equations
Problem 6.87 · medium
Solve \( \displaystyle y'' - 4y' + 3y = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = 2 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 3\right) \left(m - 1\right) = m^{2} - 4 m + 3 \]Roots: 3, 1.✓ Proved
- Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ 6 e^{x} - 4 \frac{d}{d x} 2 e^{x} + \frac{d^{2}}{d x^{2}} 2 e^{x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 2 \]y(0).✓ Proved
- \[ 2 \]y'(0).✓ Proved
Answer \( y = 2 e^{x} \)
Lines: 4 proved, 3 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 | Not checked | — | a sentence; read, not computed |
| 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 |
| 7 | ✓ 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; both initial conditions match |
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/second_order_linear, checked 2026-09-26 with SymPy 1.14.0.