**A Linear Second Order Differential Equation is of the form:**
$a(x) \frac{d^2 y}{dx^2} +b(x) \frac{dy}{dx}+c(x)y=d(x)$
Or equivalently,

$a(x) y''+b(x) y'+c(x)y=d(x)$
Where all of

$a(x),b(x),c(x),d(x)$ are functions of

$x$.

A Linear Second Order Differential Equation is called

__homogeneous__ if

$d(x)=0$. So,

$a(x) y''+b(x) y'+c(x)y=0$
And a general constant coefficient linear homogeneous, second order differential equation looks like this:

$Ay''+By'+Cy=0$
Let’s suppose that both f(x) and g(x) are solutions to the above differential equations, then so is

$y(x)=c_1 f(x)+c_2 g(x)$
Where

$c_1$ and

$c_2$ are constants

__Characteristic Equation__
The general solution to the differential equation:

$Ay''+By'+Cy=0$
Will be of the form:

$y(x)=e^{rx}$
Taking the derivatives:

$y' (x)=re^{rx}$
$y'' (x)=r^2 e^{rx}$
And inputting them into the above equation:

$Ar^2 e^{rx}+Bre^{rx}+Ce^{rx}=0$
So we will have:

$e^{rx} (Ar^2+Br+C)=0$
The equation

$Ar^2+Br+C=0$ is called the

__Characteristic Equation__. And is used to solve these sorts of questions.

Solving the quadratic we will get some values for

$r$:

Real Roots:

$r_1 \neq r_2$
Complex Roots:

$r_1,r_2=\lambda \pm \mu i$
Repeated Real Roots:

$r_1=r_2$
Though let’s deal with only real roots for now

As we will get two solutions to the characteristic equation:

$y_1 (x)=e^{r_1 x}$ $y_2 (x)=e^{r_2 x}$
So all solutions will be of the form:

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