Initial value problem y''+2y'+5y=0

Short answer

Roots of m^2+2m+5=0 are -1±2i, so y = e^{-x}(A cos2x + B sin2x). y(0)=1 gives A=1; y'(0)=5 gives -1+2B=5, B=3. Answer: y = e^{-x}(cos2x + 3 sin2x).

Solve the initial value problem y'' + 2y' + 5y = 0, given y(0) = 1, y'(0) = 5.

Equation: y″ + 2y′ + 5y = 0, y(0) = 1, y′(0) = 5.

Auxiliary equation: m² + 2m + 5 = 0, so m = (−2 ± √(4 − 20))/2 = −1 ± 2i.

The roots are complex (α ± iβ with α = −1, β = 2), so

y = e⁻ˣ (A cos 2x + B sin 2x).

y(0) = 1: A = 1.

y′ = −e⁻ˣ(A cos 2x + B sin 2x) + e⁻ˣ(−2A sin 2x + 2B cos 2x).

y′(0) = −A + 2B = 5, so −1 + 2B = 5 and B = 3.

Answer: y = e⁻ˣ (cos 2x + 3 sin 2x).

Check: y(0) = 1 ✓; y′(0) = −1 + 6 = 5 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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