# Solve Second Order Differential Equations - part 1

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

## Examples

Example 1: Solve the second order differential equation given by**y" + 2 y' - 3 y = 0**

__Solution to Example 1__

The auxiliary equation is given by

k^{2} + 2 k - 3 = 0

Factor the above quadratic equation and solve for k

(k + 3)(k - 1) = 0

k1 = -3 and k2 = 1

The general solution to the given differential equation is given by

y = A e^{k1 x} + B e^{k2 x} = A e^{-3 x} + B e^{x}

where A and B are constants.

Example 2: Solve the second order differential equation given by

**y" + 3 y' -10 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} + 3 k - 10 = 0

Solve the above quadratic equation to obtain

k1 = 2 and k2 = - 5

The general solution to the given differential equation is given by

y = A e^{2 x} + B e^{- 5 x}

where A and B are constants that may be evaluated using the initial conditions. y(0) = 1 gives

y(0) = A e^{0} + B e^{0} = A + B = 1

y'(0) = 0 gives

y'(0) = 2 A e^{0} - 5 B e^{0} = 2 A - 5 B = 0
Solve the system of equations A + B = 1
and 2 A - 5 B = 0 to obtain

A = 5 / 7 and B = 2 / 7

The solution to the given equation may be written as

y = (5 / 7) e^{2 x} + (2 / 7) e^{- 5 x}

## Exercises:

Solve the following differential equations.1. y" + 5 y' - 6 y = 0

2. y" + y' - 2 y = 0 with the initial conditions y(0) = 2 and y'(0) = 0

__Answers to Above Exercises__

1. y = A e

^{ x}+ B e

^{ -6 x}, A and B constant

2. y = (4 / 3) e

^{ x}+ (2 / 3) e

^{ -2 x}

More references on
Differential Equations

Differential Equations - Runge Kutta Method