The auxiliary equation of a second order differential equation d^{2}y / dx^{2} + b dy / dx + c y = 0 is given by
k^{2} + b k + c = 0
If b^{2}  4c is < 0, the equation has 2 complex conjugate solution of the form
k1 = r + t i and k2 = r  t i , where i is the imaginary unit.
In such case, it can be shown that the general solution to the second order differential equation may be written as follows
y = e^{r x} [ A cos x t + B sin x t ] where A and B are constants.
Example 1: Solve the second order differential equation given by
y" + y' + 2 y = 0
Solution to Example 1
The auxiliary equation is given by
k^{2} + k + 2 = 0
Solve for k to obtain 2 complex conjugate solutions
k1 = 1 / 2  i sqrt(7)
and k2 = 1 / 2 + i sqrt(7) ,
r = 1/2 (real part)
and t = sqrt(7) (imaginary part)
The general solution to the given differential equation is given by
y = e^{ x / 2} [ A cos (sqrt(7) x) + B sin (sqrt(7) x) ]
where A and B are constants.
Example 2: Solve the second order differential equation given by
y" + sqrt(3) y' + 3 y = 0
with the initial conditions y(0) = 1 and y'(0) = 0
Solution to Example 2
The auxiliary equation is given by
k^{2} + sqrt(3) k + 3 = 0
Solve the quadratic equation to obtain
k1 =  1/2 + 3 i and k2 = 1/2  3 i
The general solution to the given differential equation is given by
y = e^{(1/2) x}[ A sin 3x + B cos 3x ]
The initial condition y(0) = 1 gives
y(0) = e^{0} [ A sin 0 + B cos (0) ] = 1 which gives B = 1
y'(0) = 0 gives
y'(0) = (1/2)e^{0} [ A sin 0 + B cos 0 ] + e^{0} [ 3 A cos 0  3 B sin 0 ]
Solve the system of equations B = 1
and (1/2) B + 3 A = 0 to obtain
A = 1/6 and B = 1
The solution may be written as
y = e^{ x / 2} [ (1 / 6) sin 3 x + cos 3 x]
Exercises: Solve the following differential equations.
1. y"  y' + y = 0
2. y" + y = 0 with the initial condtions y(0) = 1 and y'(0) = 0
Answers to Above Exercises
1. y = A e^{ x / 2} [ A cos (sqrt(3) / 2) x + B sin (sqrt(3) / 2) x ] , A and B constant
2. y = cos x
More references on
Differential Equations
Differential Equations  Runge Kutta Method
