求y''+y=0的通解

设y'=u,则y''=d(y')/dx=(du/dy)*(dy/dx)=u*(du/dy),
所以，原式化为：
u*(du/dy)+y=0
udu=-ydy
积分得
(1/2)u^2=-(1/2)y^2+C1
(dy/dx)^2=C1-y^2
dy/dx=√(C1-y^2)
dy/(C1-y^2)=dx
积分得
arcsin(y/C1)=x+C2
y/C1=sin(x+C2)
∴y=C1sin(x+C2)

C1cosx+C2sinx