设g(x)在[a,b]连续,f(x)在[a,b]二阶可导,f(a)=f(b)=0,且对x(a≤x≤b)满足f’’(x)+g(x)f’(x)一f(x)=0.求证:f(x)=0(x∈[a,b]).

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问题 设g(x)在[a,b]连续,f(x)在[a,b]二阶可导,f(a)=f(b)=0,且对x(a≤x≤b)满足f’’(x)+g(x)f’(x)一f(x)=0.求证:f(x)=0(x∈[a,b]).

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答案若f(x)在[a,b]上不恒为零,则f(x)在[a,b]取正的最大值或负的最小值. 不妨设f(x0=[*]f(x)>0,则x0∈(a,b)且f’(x0)=0,f’’(x0)≤0[*] f’’(x0)+g(x0) f’(x0)一f(x0)<O与已知条件矛盾.同理,若f(x1)=[*]f(x)<0,同样得矛盾.因此f(x)≡0 ([*]x∈[a,b]).

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