作者LuisSantos (^______^)
看板trans_math
标题Re: [考古] 台大83年考古题
时间Thu Mar 8 13:10:11 2007
※ 引述《ji394zvup (想不到)》之铭言：
: f"(x)=-4f(x) ，f(0)=3，f'(0)=0，
: 试f(x)=?
: 希望有高手能解惑
: 谢谢
f"(x) = -4f(x)
f"(x) + 4f(x) = 0
f(x)的辅助方程式为 m^2 + 4 = 0 => m^2 = -4 => m = 2i , -2i
f(x)的通解为 f(x) = (c_1)(e^0)(cos2x) + (c_2)(e^0)(sin2x)
= (c_1)(cos2x) + (c_2)(sin2x)
f'(x) = (-2)(c_1)(sin2x) + (2)(c_2)(cos2x)
f(0) = 3 => c_1 = 3
f'(0) = 0 => (2)(c_2) = 0 => c_2 = 0
所以 f(x) = 3cos2x
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1F：推 ji394zvup:谢谢^^ 122.126.101.33 03/08 13:46