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Exercise 382Calculus › Differential equations › Linear equations of order twoL1Added 2 Nov 2026

Complex conjugate roots

#characteristic equation#complex roots#oscillation

Statement

Solve the differential equation
y′′+4y′+5y=0.y'' + 4y' + 5y = 0.

Hint

The characteristic equation has complex conjugate roots.

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Solution

Characteristic equation.
r2+4r+5=0.r^2 + 4r + 5 = 0.
The discriminant is Δ=16−20=−4=(2i)2\Delta = 16 - 20 = -4 = (2i)^2, so the roots are
r1,2=−4±2i2=−2±i.r_{1,2} = \frac{-4 \pm 2i}{2} = -2 \pm i.

General solution.
y(x)=e−2x(Acos⁡x+Bsin⁡x),A,B∈R.y(x) = e^{-2x}(A\cos x + B\sin x), \qquad A,B \in \mathbb{R}.

Remark. This is the oscillatory (pseudo-periodic) case.