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If a function 'f' satisfies the relation...

If a function 'f' satisfies the relation `f(x)f^('')(x)-f(x)f^(')(x) -f^(')(x)^(2)=0 AA x in R` and `f(0)=1=f^(')(0)`. Then find `f(x)`.

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The correct Answer is:
`f(x) = e^(e^(x)-1)`
We have `f(x)f^('')(x)-f^(')(x)^(2)=0`
Divide by `f(x)f^(')(x)`, we get
`(f^('')(x))/(f^(')(x))-1=(f^(')(x))/(f(x))`
Integrating both sides, we get ,
`log_(e)f^(')(x)-x=log_(e)f(x)+C`
Since, `f(0)=1=f^(')(0),C=0`
`therefore log_(e)f^(')(x)-log_(e)f(x)=x`
`rArr (log_(e)) f^(')(x)/(f(x))=e^(x)`
Integrating both sides, we get
`log_(e)f(x)=e^(x)+C`
Using `f(0)=1`, we have `C=-1`
`therefore f(x)=e^(e^(x))-1`
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