f(x)的导数=f(x)+1,且f(0)=0,求f(x)
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f(x)的导数=f(x)+1,且f(0)=0,求f(x)
f(x)的导数=f(x)+1,且f(0)=0,求f(x)
f(x)的导数=f(x)+1,且f(0)=0,求f(x)
令y=f(x)
y'=y+1
dy/dx=y+1
dy/(y+1)=dx
∫dy/(y+1)=∫dx
ln|y+1|=x+C
y+1=e^(x+C)
y=e^(x+C)-1
当x=0时 y=0
所以0=e^C-1 => e^C=1 =>C=0
所以f(x)=e^x-1
由题可知:f'(x)=f(x)+1
--->[f(x)+1]'=f(x)+1
--->(ln[f(x)+1])'=1
---->ln[f(x)+1]=x+C
由f(0)=0,代入上式,可得:ln1=0+C--->C=0
所以ln[f(x)+1]=x
f(x)+1=e^(x)
f(x)=e^x-1
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