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

## 2nd Order Differential Equation

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

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

2nd Order Differential Equations above can be rewritten as shown

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

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

$s^{2}+{\frac {b}{a}}s+{\frac {c}{a}}=0$ $s=(-\alpha \pm {\sqrt {\lambda }}$ ) t
$\alpha ={\frac {b}{2a}}$ $\beta ={\frac {c}{a}}$ $\lambda ={\sqrt {\alpha ^{2}-\beta ^{2}}}$ ### Case 1

When

$\lambda =0$ Then

$\alpha ^{2}=\beta ^{2}$ $s=e^{(}-\alpha t)$ Equation has one real roots

### Case 2

When

$\lambda >0$ Then

$\alpha ^{2}>\beta ^{2}$ $s=e^{(}-\alpha x)e^{[}\pm (\lambda x)]$ Equation has two real roots

### Case 3

When

$\lambda <0$ Then

$\alpha ^{2}<\beta ^{2}$ $s=e^{(}-\alpha t)[e^{(}\pm j\lambda t)]$ Equation has two compex roots

## Special Case

### Case 1

Differential Equation of the form

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

$s=\pm j{\sqrt {\lambda }}$ ### Case 2

Differential Equation of the form

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

$s=\pm {\sqrt {\lambda }}$ ## Summary

2nd Order Differential Equation

${\frac {d^{2}}{dx^{2}}}f(x)+{\frac {b}{a}}{\frac {d}{dx}}f(x)+{\frac {c}{a}}=0$ has roots depend on the value of $\lambda$ 1. $\lambda =0.f(x)=e^{(}-\alpha x)$ 2. $\lambda >0.f(x)=e^{(}-\alpha x)e^{(}\pm \lambda x)$ 3. $\lambda <0.f(x)=e^{(}-\alpha x)e^{(}\pm j\lambda x)$ With

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