# Solve Second Order Differential Equations - part 3

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 = er 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 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. Example 2: Solve the second order differential equation given by y" + √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 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: Solve the following differential equations. 1. y" - y' + y = 0 2. y" + y = 0 with the initial conditions y(0) = 1 and y'(0) = 0 Answers to Above Exercises 1. y = A e x / 2 [ A cos (√3 / 2) x + B sin (√3 / 2) x ] , A and B constant 2. y = cos x More references on Differential Equations Differential Equations - Runge Kutta Method