来自刘庆纲的问题
【设f(x)在[a,b]有二阶连续导数,满足方程f''(x)+x²f'(x)-2f(x)=0,证若f(a)=f(b)=0,则f(x)在[a,b]恒为0】
设f(x)在[a,b]有二阶连续导数,满足方程f''(x)+x²f'(x)-2f(x)=0,证若f(a)=f(b)=0,则f(x)在[a,b]恒为0
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2020-12-24 11:56
【设f(x)在[a,b]有二阶连续导数,满足方程f''(x)+x²f'(x)-2f(x)=0,证若f(a)=f(b)=0,则f(x)在[a,b]恒为0】
设f(x)在[a,b]有二阶连续导数,满足方程f''(x)+x²f'(x)-2f(x)=0,证若f(a)=f(b)=0,则f(x)在[a,b]恒为0
反证法.假设f(x)在[a,b]上不恒为零,不妨设c∈(a,b),f(c)>0且是f(x)在[a,b]上的最大值,于是f'(c)=0,且f''(c)