Arithmetic Course/Differential Equation/Second Order Equation

Second Order Differential Equation

Second Ordered Differetial Equation has the general form

$A \frac{d^2 f(x)}{dx^2} + B \frac{d f(x)}{dx} + C = 0$

Which can be expressed as

$\frac{d^2 f(x)}{dx^2} + \frac{B}{A} \frac{d f(x)}{dx} + \frac{C}{A} = 0$

Solving 2nd Ordered Differential Equation

$A \frac{d^2 f(x)}{dx^2} + B \frac{d f(x)}{dx} + C = 0$

$\frac{d^2 f(x)}{dx^2} + \frac{B}{A} \frac{d f(x)}{dx} + \frac{C}{A} = 0$

$s^2 + \frac{B}{A} s + \frac{C}{A} = 0$

$s = (-\alpha \pm \sqrt{\alpha^2 - \beta^2}) x$

$s = (-\alpha \pm \lambda) x$

Case 1

$\lambda = 0$

$\alpha^2 = \lambda^2$

$s = -\alpha x$

$f(x) = e^(-\alpha x)$

Case 2

$\lambda > 0$

$\alpha^2 > \lambda^2$

$s = -\alpha x \pm \lambda x$

$f(x) = e^(\alpha x) [e^(-\alpha x) + e^(-\alpha x)]$

$f(x) = A e^(\alpha x) Cos \lambda x$

$A = \frac{1}{2} e^(\alpha x)$

Case 3

$\lambda < 0$

$\alpha^2 < \lambda^2$

$s = -\alpha x \pm j \lambda x$

$f(x) = e^(-\alpha x) [e^(\alpha x) + e^(-j\alpha x)]$

$f(x) = A e^(\alpha x) Sin \lambda x$

$A = \frac{1}{2j} e^(\alpha x)$

Last modified on 6 May 2011, at 16:14