ZoroP said:
well, there maybe some misunderstandings here between us. Like when x approach to 0, what I know is siny = y, so I change y''+g*siny\L = 0 to y'' + g*y\L=0. And your formula is not what I used. Whatever, thanks anyway~
"when x approach to 0, what I know is siny = y". No, you don't know that- it makes no sense. Perhaps you meant "when x approach 0, sin x= x" but that still is not true. For values of x very close to 0, sin x is very close to x. That results in \lim_{x\rightarrow 0} sin x/x= 1
What you originally wrote was "y''+g*siny\L = 0 but then how to find the linear approximation about y'' + g*y\L=0". As both you and I have said now, for small y, sin y is close to y so approximately, replacing sin y by y, y"+ gy/L= 0. That's what you wanted.
Now, do you know how to find the characteristic equation for y"+ (g/L)y= 0? What are its roots?
You need to know that if a\pm ib are roots of the characteristic equation of a "linear homogenouse differential equation with constant coefficients", then
y(x)= e^{at}(C_1 cos(bt)+ C_2 sin(bt)
is the general solution to the differential equation.
We could write
y(t)= D_1e^{(a+ bi)t}+ D_2e^{(a+bi)t}= e^{at}(D_1e^{bit}+ D_2e^{-bit}
but e^{bit}= cos(bt)+ i sin(bt) and e^{-bit}= cos(bt)- i sin(bt)
We can combine those and then absorb the "i" into the constant. We would expect that if the original problem involved only real numbers, the solution will involve only real numbers.