# 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
k2 + 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 ek1 x + B ek2 x = A e-3 x + B ex
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
k2 + 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 e2 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 e0 + B e0 = A + B = 1
y'(0) = 0 gives
y'(0) = 2 A e0 - 5 B e0 = 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) e2 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