# Homogeneous second-order linear ordinary differential equation with constant coefficients

## Equation

Governing equation:

 $y'' + a y' + b y = 0$ (1)

where $a$ and $b$ are real constants.

## Solution

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

 $f(\lambda) = \lambda^2 + a \lambda + b = 0$ (2)

Then

 (i) If $\lambda_1 \ne \lambda_2$ real, then $y = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_1 x}$ (3) (ii) If $\lambda = \lambda_1 = \lambda_2$, then $y = (C_1 x + C_2) e^{\lambda x}$ (iii) If $\lambda_1 = \alpha + i \beta$, $\lambda_1 = \alpha - i \beta$ ($\beta \ne 0$), then $y = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin (\beta x)) = e^{\alpha x} C \cos(\beta x + \theta)$

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.

To Do: derivation of solution for second order linear ordinary differential equation - clean up stuff. (who: user:ok; priority: 100; hours: 0) (all) (cat)