# 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 $$y'(0) = 0$$
1. $$y = A e^x + B e^{-6 x}$$, $$A$$ and $$B$$ constant
2. $$y = \left(\frac{4}{3}\right) e^x + \left(\frac{2}{3}\right) e^{-2 x}$$