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.
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 condtions 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

