Solve Second Order Differential Equations
part 3

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 + A sin x t ] where A and B are constants.

Example 1: Solve the second order differential equation given by

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 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

Solution to Example 2

The auxiliary equation is given by

k² + 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) = e⁰ [ A sin 0 + B cos (0) ] = 1 which gives B = 1

y'(0) = 0 gives

y'(0) = -(1/2)e⁰ [ A sin 0 + B cos 0 ] + e⁰ [ 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