Solve Second Order Differential Equations - Two distinct real solutions

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 with Solutions

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
\[ (k + 3)(k - 1) = 0 \]
Solve for \( k \)
\[ k_1 = -3 \; \text{and} \; k_2 = 1 \]
The general solution to the given differential equation is given by
\[ y = A e^{k_1 x} + B e^{k_2 x} = A e^{-3 x} + B e^x \]
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
\( k_1 = 2 \) and \( k_2 = -5 \)
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
\( A = \frac{5}{7} \) and \( B = \frac{2}{7} \)
The solution to the given equation may be written as
\[ y = \left(\frac{5}{7}\right) e^{2 x} + \left(\frac{2}{7}\right) e^{- 5 x} \]

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. \( y = \left(\frac{4}{3}\right) e^x + \left(\frac{2}{3}\right) e^{-2 x} \)