设P(x)在[0,+∞)连续且为负值,y=y(x)在[0,+∞)连续,在(0,+∞)满足y’+P(x)y>0且y(0)≥0,求证:y(x)在[0,+∞)单调增加.

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问题 设P(x)在[0,+∞)连续且为负值,y=y(x)在[0,+∞)连续,在(0,+∞)满足y’+P(x)y>0且y(0)≥0,求证:y(x)在[0,+∞)单调增加.

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答案由y’+P(x)y>(x>0)[*][e0xP(t)dty(x)]’>0((x>0),又e0xP(t)dty(x)在[0,+∞)连续,[*]e0xP(t)dty(x)在[0,+∞]单调[*]e0xP(t)dty(x)>[e0xP(t)dty(x)|x0=y(0)≥0[*]y(x)>0(x≥0)[*]y’(x)>-P(x)y(x)>0(x>0)[*]y(x)在[0,+∞)单调增加.

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