Calculus Course/Differential Equations/2nd Order Differential Equations

2nd Order Differential Equation

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

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

Characteristic Equation

2nd Order Differential Equations above can be rewritten as shown

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

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

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

$s^2=-\lambda$

Roots of equation

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

Case 2

Differential Equation of the form

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

$s^2=\lambda$

Roots of equation

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

Summary

2nd Order Differential Equation

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

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

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

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

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

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