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.
Consider the differential equation
\[ y'' + a y' + b y = 0 \]We form the characteristic (auxiliary) equation
\[ k^2 + a k + b = 0 \]If this quadratic equation has two distinct real roots \( k_1 \neq k_2 \), then the general solution is
\[ y = A e^{k_1 x} + B e^{k_2 x} \]where \( A \) and \( B \) are constants.
Solve:
\[ y'' + 2y' - 3y = 0 \]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:
\[ y = A e^{-3x} + B e^{x} \]Solve:
\[ y'' + 3y' - 10y = 0 \]with
\[ y(0)=1,\quad y'(0)=0 \]Auxiliary equation:
\[ k^2+3k-10=0 \]Roots:
\[ k_1=2,\quad k_2=-5 \]General solution:
\[ y=Ae^{2x}+Be^{-5x} \]Apply initial conditions:
\[ A+B=1 \] \[ 2A-5B=0 \]Solving gives:
\[ A=\frac{5}{7},\quad B=\frac{2}{7} \]Final solution:
\[ y=\frac{5}{7}e^{2x}+\frac{2}{7}e^{-5x} \]