2. 考研
  3. 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)－f(x)=0．求证：f(x)≡0 (x∈[a，b])．

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)－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)－f(x)=0．求证：f(x)≡0  (x∈[a，b])．
选项
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)=[*]＞0，则x0∈(a，b)且f’(x0)=0，f"(x0)≤0 => f"(x0)+g(x0)f’(x0)－f(x0)＜0与已知条件矛盾．同理，若f(x1)=[*]＜0，同样得矛盾．因此f(x)≡0 ([*]x∈[a，b])．
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考研数学二

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