## Linear Second-Order Equations with Constant Coefficients

Homogeneous linear differential equations of the form

L(y) = a(x)y'' + b(x)y' + c(x)y
= 0

where a(x), b(x), and c(x) are constants (i.e., a(x) = a, b(x) = b, and c(x) = c) may be rewritten as:

ay'' + by' + cy = 0

Letting y = e

^{rx}, it follows that y' = re

^{rx }and y'' = r

^{2}e

^{rx }. After substiting these variables into the above equation, results in the following:

ar

e

^{2}e^{rx}+ bre^{rx}+ ce^{rx}= 0e

^{rx}(ar^{2}+ br + c) = 0Therefore, e

^{rx}is a solution if and only if r is a root of the characteristic equation:ar

Two distinct conjugate complex roots r ± vi yielding the following solution:

y = e

Example:

Solution:

^{2}+ br + c = 0The roots of the
characteristic equation are provided by the quadratic formula:

y = c

Two real and identical roots r

y = c

1) For the
discriminant b

^{2}− 4ac > 0:_{1}and r

_{2}yielding the following solution:

y = c

_{1}e

^{r1x}+ c

_{2}e

^{r2x }

2) For the
discriminant b

^{2}− 4ac = 0:Two real and identical roots r

_{1}and r

_{2}= r yielding the following solution:

y = c

_{1}e

^{rx}+ c

_{2}xe

^{rx}

3) For the
discriminant b

^{2}− 4ac < 0:Two distinct conjugate complex roots r ± vi yielding the following solution:

y = e

^{rx}(c

_{1}cos(vx) + c

_{2}sin(vx))

Example:

Solve
y'' − 5y' − 6y
= 0

Solution:

As a = 1, b = −5, c = −6, resulting
characteristic equation is:

r

Roots of above equation may be determined to be r

Therefore, solutions of the differential equation are e

y(x) = c

r

^{2}− 5r − 6 = 0Roots of above equation may be determined to be r

_{1}= −1 and r_{2}= 6Therefore, solutions of the differential equation are e

^{-x}and e^{6x}with the general solution provied by:y(x) = c

_{1}e^{-x }+ c_{2}e^{6x}^{ }