设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x).

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/30 02:26:50
设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x).

设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x).
设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x).

设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x).
答:f(x) = 2sinx + cosx
f(x) = 1 + 2x + ∫(0~x) tf(t) dt - x∫(0~x) f(t) dt ...(1)
f'(x) = 2 + xf(x) - [∫(0~x) f(t) dt + xf(x)]
f'(x) = 2 - ∫(0~x) f(t) dt
f''(x) = -f(x)
f''(x) + f(x) = 0 ...(2)
特征方程:r² + 1 = 0 => r = ±i
f(x) = Asinx + Bcosx,A、B为任意常数
由(1):f(0) = 1
=> f(0) = Asin(0) + Bcos(0) = B
=> B = 1
f(x) = Asinx + cosx,代入(1):
Asinx + cosx = 1 + 2x + ∫(0~x) (t - x)(Asint + cost) dt
Asinx + cosx = 1 + 2x + Asinx + cosx - Ax - 1
=> A = 2
所以f(x) = 2sinx + cosx

设函数f(x)在(-∞,+∞)可导,且满足f(0)=1,f'(x)=f(x),证明f(x)=e^x 设f(x)为可导函数,且满足f(x)=∫(上限X下线1)f(t)/tdt+(x-1)e^x求f(x) 设函数f(x)可导,且满足xf'(x)=f'(-x)+1,f(0)=0,求函数f(x)的极值 设函数可导,且满足xf'(x)=f'(-x)+1,f(0)=0 求f'(x) 求f(x)的极限 设函数f(x)可导,且满足f(0)=0,又f'(x)单调减少.证明对x∈(0,1),有f(1)x 设函数f(x)可导,且满足f(x)-∫(上限为x,下限为0)f(t)dt=e^x,求f(x) 需要详解, 设函数f(x)可导,且满足f(x)=x²+∫(0~x)f(t)dt 求f(x)如题 8、设f(x)为可导函数,且满足∫0到x f(t)t^2 dt=f(x)+3x 求f(x) 设函数f(x)可导,且满足f(x)=x^2+∫0~x f(t)dt,求f(x) 8、设f(x)为可导函数,且满足∫0到x f(t)t^2 dt=f(x)+3x 求f(x) 设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函数f(x). 设函数f(x)可微,且满足§[2f(t)-1]dt=f(x)-1,求f(x) 设f(x)为可导函数,且满足lim[f(1)+f(1-x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率 设f(x)为可导函数,且满足lim[f(1)-f(1-x)]/2x=-2,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率 设函数f(x) 可导,且f(0)=1 ,f'(-lnx)=x ,则f(1)= 设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率设f(x)为可导函数,且满足lim[f(1)-f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率 设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率设f(x)为可导函数,且满足lim[f(1)-f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率 设定义在R上的可导函数f(x)满足f(1+x)=f(1-x),且当x∈【-∞,1】时(x-1)f ’(x)