A tutorial on how to solve second order differential equations with auxiliary equation having 2 distinct complex solutions. Examples with detailed solutions are included.

The auxiliary equation of a second order differential equation d²y / dx² + b dy / dx + c y = 0 is given by
k² + b k + c = 0
If b² - 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.

Examples

Solution to Example 1
The auxiliary equation is given by
k² + k + 2 = 0
Solve for k to obtain 2 complex conjugate solutions
k1 = -1 / 2 - i √7 / 2
and k2 = -1 / 2 + i √7 / 2,
r = -1/2 (real part)
and t = √7 /2 (imaginary part)
The general solution to the given differential equation is given by
y = e^{- x / 2} [ A cos ((√7 /2) x) + B sin ((√7/2) x) ]
where A and B are constants.

Solution to Example 2
The auxiliary equation is given by
k² + √3 k + 3 = 0
Solve the quadratic equation to obtain
k1 = - √3/2 + 3/2 i and k2 = - √3/2 - 3/2 i
The general solution to the given differential equation is given by
y = e^{-(√3/2) x}[ A sin (3/2)x + B cos (3/2)x ]
The initial condition y(0) = 1 gives
y(0) = e⁰ [ A sin 0 + B cos (0) ] = 1 which gives B = 1
y'(0) = 0 gives
y'(0) = -(√3/2)e⁰ [ A sin 0 + B cos 0 ] + e⁰ [ (3/2) A cos 0 - (3/2) B sin 0 ]
Solve the system of equations B = 1 and -(√3/2) B + (3/2) A = 0 to obtain
A = √3/3 and B = 1
The solution may be written as
y = e^{-(√3/2) x} [ (√3/3) sin (3/2) x + cos (3/2) x]

Exercises

More references on Differential Equations
Differential Equations - Runge Kutta Method