A tutorial on how to solve second order differential equations with auxiliary equation having 2 distinct real solutions. Examples with detailed solutions are included.
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.
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
\[ y = \left(\frac{5}{7}\right) e^{2 x} + \left(\frac{2}{7}\right) e^{- 5 x} \]