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

$\alpha ={\frac {b}{a}}$

$\beta ={\frac {c}{a}}$

$\lambda ={\sqrt {\alpha ^{2}-\beta ^{2}}}$