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)`.

Options

A. `(1)/(sqrt(e ))`
B. `(1)/(e )`
C. `sqrt(e )`
D. e

Answer

B

Explanation

`f(x)f''(x)-f(x)f'(x)=(f'(x))^(2)`
` :. (f''(x)-f'(x))/(f'(x))=(f'(x))/(f(x))`
`implies(f''(x))/(f'(x))-1=(f'(x))/(f(x))`
On integrating both sides w.r.t. x, we get
`log_(e)(f'(x))-x=log_(e)f(x)+c`
`f(0)=f'(0)=1`
` :. 0-0=0+c`
` :. c=0`
` :. log_(e)(f'(x))-x=log_(e)f(x)`
`implieslog_(e)(f'(x))-log_(e)f(x)=x`
`implies (f'(x))/(f(x))=e^(x)`
On integrating both sides w.r.t. x. we get
` :. log_(e)f(x)=e^(x)+c`
Putting `x=0`, we get `c= -1`
` :. log_(e)f(x)=e^(x)-1`
` :. f(x)=e^(e^(x)-1)`