# Homogeneous linear second order differential equations

##### Intros

###### Lessons

##### Examples

###### Lessons

**Solving Homogeneous Linear Second Order Differential Equations**

Find some general solutions to the following constant coefficient homogeneous linear second order differential equation:

$2y''+5y'-3y=0$- Using the initial conditions find a particular solution to the following differential equation:

$y''+3y'-4y=0$ $y(0)=3, y' (0)=-2$

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###### Topic Notes

**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:

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

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