# Characteristic equation with real distinct roots

### Characteristic equation with real distinct roots

#### Lessons

In the previous section we came up with a method to solve linear homogeneous constant coefficient second order differential equations:

$Ay''+By'+Cy=0$

By using the characteristic equation:By using the characteristic equation:

$Ar^2+Br+C=0$

And solving this quadratic will yield two roots, $r_1,r_2$. Let’s suppose that both $r_1$ and $r_2$ are distinct and real.

So the solution will be:

$y_1 (x)=e^{r_1 x}$
$y_2 (x)=e^{r_2 x}$

Or in full generality:

$y(x)=c_1 e^{r_1 x}+c_2 e^{r_2 x}$

This is the general solution. We can find a particular solution with initial parameters

$y(x_0 )=y_0, y' (x_1 )=y_1$.
• Introduction
Using the Characteristic Equation with Real Distinct Roots

• 1.
Using the Characteristic Equation with Real Distinct Roots
Find the particular solution to the following differential equation:

$y''-9y=0$

With initial values $y(0)=2, y' (0)=1$

• 2.
Find the particular solution to the following differential equation:

$6y''+8y'-8y=0$

With initial values $y(2)=1, y' (2)=4$

• 3.
Find the particular solution to the following differential equation:

$y''-5y'+3y=0$

With initial values $y(0)=0, y' (0)=1$