【转载请注明出处】http://www.cnblogs.com/mashiqi
2015/09/08
Today we focus on the following equation:
$$u''=au, extrm{where} (a > 0)$$
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 $f'-2sqrt{a}f=C Rightarrow (f+frac{C}{2sqrt{a}})'=2sqrt{a}(f+frac{C}{2sqrt{a}})$, so $ln|f+frac{C}{2sqrt{a}}|=2sqrt{a}x+C' Rightarrow f = C'e^{2sqrt{a}x} - frac{C}{2sqrt{a}}$. Finally we get:
$$u(x) = (C'e^{2sqrt{a}x} - frac{C}{2sqrt{a}}) e^{-sqrt{a}x} = A e^{sqrt{a}x} + B e^{-sqrt{a}x}$$