This tutorial explains how to solve homogeneous second order linear differential equations with constant coefficients when the characteristic (auxiliary) equation has two distinct real roots. Step-by-step examples and exercises are included.
General Method
Consider the differential equation:
\[ y'' + ay' + by = 0 \]We form the characteristic (auxiliary) equation:
\[ k^2 + ak + b = 0 \]If this quadratic equation has two distinct real roots \( k_1 \neq k_2 \), then the general solution is:
where \( A \) and \( B \) are arbitrary constants.
Worked Examples
Step-by-Step Solved Examples
Example 1
Solve:
\[ y'' + 2y' - 3y = 0 \]Solution:
The auxiliary equation is:
\[ k^2 + 2k - 3 = 0 \]Factor:
\[ (k + 3)(k - 1) = 0 \]Roots:
\[ k_1 = -3, \quad k_2 = 1 \]General solution:
\[ y = Ae^{-3x} + Be^{x} \]Example 2 (With Initial Conditions)
Solve:
\[ y'' + 3y' - 10y = 0 \]subject to the initial conditions:
\[ y(0) = 1, \quad y'(0) = 0 \]Solution:
Auxiliary equation:
\[ k^2 + 3k - 10 = 0 \implies (k + 5)(k - 2) = 0 \]Roots:
\[ k_1 = 2, \quad k_2 = -5 \]General solution:
\[ y = Ae^{2x} + Be^{-5x} \]Apply initial conditions to find \( A \) and \( B \):
1) \( y(0) = A e^0 + B e^0 = A + B = 1 \)
Compute derivative \( y'(x) = 2Ae^{2x} - 5Be^{-5x} \):
2) \( y'(0) = 2A - 5B = 0 \)
Solving the system of equations gives:
\[ A = \frac{5}{7}, \quad B = \frac{2}{7} \]Final solution:
\[ y = \frac{5}{7}e^{2x} + \frac{2}{7}e^{-5x} \]Practice Exercises and Answers
Practice Problems & Solutions
Solve the following differential equations:
- \( y'' + 5y' - 6y = 0 \)
- \( y'' + y' - 2y = 0 \), with \( y(0) = 2 \), \( y'(0) = 0 \)
Answers:
- \( y = Ae^{x} + Be^{-6x} \)
- \( y = \dfrac{4}{3}e^{x} + \dfrac{2}{3}e^{-2x} \)