# Calculus Course/Differential Equations/2nd Order Differential Equations

## 2nd Order Differential Equation

2nd Order Differential Equation is an equation that has the general form

${\displaystyle a{\frac {d^{2}}{dx^{2}}}f(x)+b{\frac {d}{dx}}f(x)+c=0}$

## Characteristic Equation

2nd Order Differential Equations above can be rewritten as shown

${\displaystyle {\frac {d^{2}}{dx^{2}}}f(x)+{\frac {b}{a}}{\frac {d}{dx}}f(x)+{\frac {c}{a}}=0}$

Let

${\displaystyle s={\frac {d}{dx}}}$

Then

${\displaystyle s^{2}+{\frac {b}{a}}s+{\frac {c}{a}}=0}$
${\displaystyle s=(-\alpha \pm {\sqrt {\lambda }}}$) t
${\displaystyle \alpha ={\frac {b}{2a}}}$
${\displaystyle \beta ={\frac {c}{a}}}$
${\displaystyle \lambda ={\sqrt {\alpha ^{2}-\beta ^{2}}}}$

### Case 1

When

${\displaystyle \lambda =0}$

Then

${\displaystyle \alpha ^{2}=\beta ^{2}}$
${\displaystyle s=e^{(}-\alpha t)}$

Equation has one real roots

### Case 2

When

${\displaystyle \lambda >0}$

Then

${\displaystyle \alpha ^{2}>\beta ^{2}}$
${\displaystyle s=e^{(}-\alpha x)e^{[}\pm (\lambda x)]}$

Equation has two real roots

### Case 3

When

${\displaystyle \lambda <0}$

Then

${\displaystyle \alpha ^{2}<\beta ^{2}}$
${\displaystyle s=e^{(}-\alpha t)[e^{(}\pm j\lambda t)]}$

Equation has two compex roots

## Special Case

### Case 1

Differential Equation of the form

${\displaystyle {\frac {d^{2}f(t)}{dt^{2}}}+\lambda =0}$
${\displaystyle s^{2}=-\lambda }$

Roots of equation

${\displaystyle s=\pm j{\sqrt {\lambda }}}$

### Case 2

Differential Equation of the form

${\displaystyle {\frac {d^{2}f(t)}{dt^{2}}}-\lambda =0}$
${\displaystyle s^{2}=\lambda }$

Roots of equation

${\displaystyle s=\pm {\sqrt {\lambda }}}$

## Summary

2nd Order Differential Equation

${\displaystyle {\frac {d^{2}}{dx^{2}}}f(x)+{\frac {b}{a}}{\frac {d}{dx}}f(x)+{\frac {c}{a}}=0}$

has roots depend on the value of ${\displaystyle \lambda }$

1. ${\displaystyle \lambda =0.f(x)=e^{(}-\alpha x)}$
2. ${\displaystyle \lambda >0.f(x)=e^{(}-\alpha x)e^{(}\pm \lambda x)}$
3. ${\displaystyle \lambda <0.f(x)=e^{(}-\alpha x)e^{(}\pm j\lambda x)}$

With

${\displaystyle \alpha ={\frac {b}{a}}}$
${\displaystyle \beta ={\frac {c}{a}}}$
${\displaystyle \lambda ={\sqrt {\alpha ^{2}-\beta ^{2}}}}$