Homogeneous second-order linear ordinary differential equation with constant coefficients

Equation
Governing equation:

where a and b are real constants.

Solution
Let \lambda_1 and \lambda_2 be the roots of the characteristic equation

Then

References: Bubenik 1997.

Derivation
The solution is expected in the form of y = e^{\lambda x} and the goal is to determine \lambda.

Express derivatives of y:

y = e^{\lambda x}

y' = \lambda e^{\lambda x}

y'' = \lambda^2 e^{\lambda x}

Substituting back to the original equation yields:

\lambda^2 e^{\lambda x} + a \lambda e^{\lambda x} + b e^{\lambda x} = 0

e^{\lambda x} (\lambda^2 + a \lambda + b) = 0

Since e^{\lambda x} \ne 0, then the following characteristic equation must be satisfied:

\lambda^2 + a \lambda + b = 0

The solution of the quadratic characteristic equation is:

\lambda_{1,2} = \frac{-a \pm \sqrt{a^2-4b}}{2}

Two Real Roots
If \sqrt{a^2-4b} > 0, then the characteristic equation has two real roots and the solution is:

y = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x}

One Real Root
If \sqrt{a^2-4b} = 0, then the characteristic equation has a single real root and the solution is:

y = C_1 e^{\lambda_1 x} + C_2 x e^{\lambda_1 x}

No Real Root
If \sqrt{a^2-4b} < 0, then the characteristic equation has a two complex roots.

Usage in Structural Engineering

 * Single degree of freedom damped free vibrations