F(x)是f(x)的原函数，G(x)是1/f(x)的原函数F(x)*G(x)=-1,f(0)=1求f(x)

Question

F(x)是f(x)的原函数，G(x)是1/f(x)的原函数F(x)*G(x)=-1,f(0)=1求f(x)

Answer 1

F(x)*G(x)=-1 所以两边关于x求导
F'(x)*G(x)+F(x)*G'(x)=0
F(x)是f(x)的原函数，G(x)是1/f(x)的原函数 所以 f(x)*G(x)=-F(x)/f(x)
两边乘上F(x)f(x)得
[F(x)]^2=[f(x)]^2, 即 f(x)=±F(x)=F'(x) ∴±1=F'(x)/F(x) ∴±1=[lnF(x)]'
所以lnF(x)=±x+c ，c为常数∴F(x)=e^[±x+c],f(x)=F'(x)=±e^[±x+c]
∵f(0)=1∴1=±e^c, 所以c=0 ,f(x)=e^x或-e^(-x)

Answer 2

F'(x)= f(x)
G'(x) = 1/f(x)
F(x).G(x) = -1

f(0)=1
------------------
F'(x)G(x) +F(x).G'(x) = 0
f(x).G(x) + F(x)/f(x) = 0
f(x)/[-F(x)] + F(x)/f(x) = 0
F(x) = f(x)
=>
f(x) = e^x +C
f(0)= 1
=> C=0
ie
f(x) = e^x