# Ordinary Differential Equations/Homogeneous second order equations

## Introduction

The general form of $2^{\text{nd}}$  order equation is

{\begin{aligned}y''=f(t,y,y').\end{aligned}}

We call them linear non-homogeneous if the equation can be written in the form
$y''+p(t)y'+q(t)y=g(t)$

and linear homogeneous if, in addition to being linear non-homogeneous, $g(t)=0$
$y''+p(t)y'+q(t)y=0.$

The method of characteristic equations is for homogeneous equations and the methods of undetermined coefficients and of variation of parameters for homogeneous equations.

## Method 1: Characteristic equation

If the equation is linear homogeneous and further $p(t),q(t)$  are constant, then the equation is referred to as a constant-coefficients equation:

$ay''+by'+cy=0$

and we can apply the method of characteristic equations to solve such an equation. Note that $a$  is assumed to be non-zero since we are working with a second order equation.

### Method formal steps

1. We assume that the solution is of the form $y(t)=e^{rt}$  (this is called making an ansatz). This gives
$(ar^{2}+br+c)e^{rt}=0\Longrightarrow \,ar^{2}+br+c=0,$

which equation is called the characteristic equation.
2. So to solve the above ODE, it suffices to find the two roots $r_{1},r_{2}$ .
3. Then the general solution is of the form:
$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}.$

### Example-presenting the method

Consider a mass $m$  hanging at rest on the end of a vertical spring of length $l$ , spring constant $k$  and damping constant $\gamma$ . Let $u\left(t\right)$  denote the displacement, in units of feet, from the equilibrium position. Note that since $u(t)$  represents the amount of displacement from the spring's equilibrium position (the position obtained when the downward force of gravity is matched by the will of the spring to not allow the mass to stretch the spring further) then $u(t)$  should increase downward. Then by Newton's Third Law one can obtain the equation

$mu''(t)+\gamma u'(t)+ku(t)=F(t),$

where $F\left(t\right)$  is any external force, which for simplicity we will assume to be zero.

1. First we obtain the characteristic equation:
$mr^{2}+\gamma r+k=0.$

2. Suppose that $m=1{\text{lb}},\gamma =5{\text{lb}}/{\text{ft}}/{\text{s}}$  and $k=6{\text{lb}}/{\text{ft}}$  then we obtain the roots $r_{1}=-2$ , $r_{2}=-3$ .
3. Therefore, the general solution will be
$u(t)=c_{1}e^{-2t}+c_{2}e^{-3t}.$

4. Further if $u(0)=0,\,u'(0)=1$  we obtain $c_{1}=1,c_{2}=-1$ :
$u(t)=e^{-2t}-e^{-3t}.$

### Examples

• Consider the IVP

$4y''-y=0,\,\,y(-2)=1,\,\,y'(-2)=-1.$

1. We obtain the characteristic equation $4r^{2}-1=0\Rightarrow r=\pm {\frac {1}{2}}$  and so the general solution will be
$y(t)=c_{1}e^{\frac {t}{2}}+c_{2}e^{-{\frac {t}{2}}}.$

2. Using the initial conditions we obtain:
$1=c_{1}e^{-1}+c_{2}e\,{\text{ and }}\,-1={\frac {1}{2}}(c_{1}e^{-1}-c_{2}e).$

3. Solving these two equations gives: $c_{1}={\frac {-1}{2}}e,c_{2}={\frac {3}{2}}e^{-1}$  and so the solution for our IVP is:
$y\left(t\right)=-{\frac {1}{2}}e^{1+{\frac {t}{2}}}+{\frac {3}{2}}e^{-{\frac {t}{2}}-1}.$

4. Therefore, as $t\to +\infty$  we obtain $y\to -\infty$ .

• Consider the IVP

$y''+5y'+6y=0,\,\,y\left(0\right)=2,\,\,y'\left(0\right)=\beta$

1. The characteristic equation is $r^{2}+5r+6=0\Rightarrow r=-2,-3$  and so the general solution will be:
$y(t)=c_{1}e^{-2t}+c_{2}e^{-3t}$

2. Using the initial conditions we obtain:
$2=c_{1}+c_{2}\,{\text{ and }}\,\beta =-2c_{1}-3c_{2}.$

3. Solving these two equations gives: $c_{1}=(6+\beta ),\,c_{2}=-\left(4+\beta \right)$  and so the solution for our IVP is:
$y(t)=(6+\beta )e^{-2t}-(4+\beta )e^{-3t}.$

4. Therefore, as $t\to +\infty$  we obtain $y\to 0$ .