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)一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)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])．
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＝[*]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])．
解析
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考研数学一

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