常微分方程的一个小推导

【转载请注明出处】http://www.cnblogs.com/mashiqi

2015/09/08

Today we focus on the following equation:
$$u''=au, \textrm{where} (a > 0)$$
Due to $a > 0$, $(u'-\sqrt{a}u)' + \sqrt{a}(u'-\sqrt{a}u)=0 \Rightarrow \frac{d(u'-\sqrt{a}u)}{u'-\sqrt{a}u}= -\sqrt{a}dx$. Therefore $u'-\sqrt{a}u = C e^{-\sqrt{a}x}$. This is a first-order linear ordinary differential equation.

Let $u(x) = f(x) e^{-\sqrt{a}x}$, we can get $f'-2\sqrt{a}f=C \Rightarrow (f+\frac{C}{2\sqrt{a}})'=2\sqrt{a}(f+\frac{C}{2\sqrt{a}})$, so $\ln|f+\frac{C}{2\sqrt{a}}|=2\sqrt{a}x+C' \Rightarrow f = C'e^{2\sqrt{a}x} - \frac{C}{2\sqrt{a}}$. Finally we get:
$$u(x) = (C'e^{2\sqrt{a}x} - \frac{C}{2\sqrt{a}}) e^{-\sqrt{a}x} = A e^{\sqrt{a}x} + B e^{-\sqrt{a}x}$$