作者profyang (prof)
看板Physics
标题Re: [问题] 应该算是简谐运动立论的基础吧,可是课ꔠ…
时间Wed Nov 14 19:19:01 2007
※ 引述《bisconect (随便你叫)》之铭言：
: if
: 1) f(0) = g(0)
: 2) f'(0) = g'(0)
: 3) for any t, f''(t) = h(f(t)) and g''(t) = h(g(t))
: then
: for any t, f(t) = g(t)
f"-g"=0; ∫(f"-g")dt=∫0dt=0=f'-g'+C1 ==>f'(0)=g'(0) ==>C1=0 ==>f'-g'=0
同理可证f-g=0 ==>f=g
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