设g(x)在[a,b]上连续,且f(x)在[a,b]上满足f"(x)+g(x)f’(x)一f(x)=0,又f(a)=f(b)=0,证明:f(x)在[a,b]上恒为零.

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问题 设g(x)在[a,b]上连续,且f(x)在[a,b]上满足f"(x)+g(x)f’(x)一f(x)=0,又f(a)=f(b)=0,证明:f(x)在[a,b]上恒为零.

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答案设f(x)在区间[a,b]上不恒为零,不妨设存在x0∈(a,b),使得f(x0)>0,则f(x)在(a,b)内取到最大值,即存在c∈(a,b),使得f(c)=M>0,且f’(c)=0,代入得f"(c)=f(c)=M>0,则x=c为极小值点,矛盾,即f(x)≤0,同理可证明f(x)≥0,故f(x)≡0(a≤x≤b).

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