How do I find this $f(x)$? I can't seem to evaluate the $f(x)$ for this problem.
Question: If a function satisfies $f(x) f"(x)-f(x)f'(x)=(f'(x))^2$ such that $x\in R$ and $f(0)=f'(0)=1$, then identify which of the following statement(s) is (are) corect?  
Statements:
$1)$Equation $f(x)=e^x$ has two solutions.  
$2)$Equation $f(x)=e^x$ has exactly one solution.  
$3)$$\displaystyle \lim_{x\rightarrow 0} \dfrac{f(x)-1}{x}=1$
$4)\displaystyle\lim_{x\rightarrow-\infty} f(x)=\dfrac{1}{e}$  
Okay, Obviously I've observed that (3) is already correct but can't really figure out $f(x)$ for the rest.
I tried applying by parts to $$f^"(x)=\frac{f'(x)(f'(x)+f(x))}{f(x)}$$ but no fruitful result. Please help.
 A: Rewrite this as : 
$$\frac{f(x)f''(x)-\big(f'(x)\big)^2}{\big(f(x)\big)^2}= \frac{f'(x)}{f(x)}$$
Do you see something?

$$d\Bigg( \frac{f'(x)}{f(x)}\Bigg) =  \frac{f'(x)}{f(x)}$$
  $$\implies   \frac{f'(x)}{f(x)}= e^x $$
  This is simple linear ODE, on solving you'll get : 
  $$f(x)= c \cdot e^{-e^x}$$
  Using $f(0)=1$;
  $$\boxed{f(x)= \frac{1}{e^{e^x-1}}}$$

Now you may proceed.
A: Let's solve for f(x) first.
Let $y = f(x)$. The original equation becomes:
$$ yy" -y'^2 = yy'$$
Divide both sides by y, assuming $y(x) \not= 0$:
$$ \frac{yy"-y'^2}{y^2} = \frac{y'}{y} $$
$$ \Longleftrightarrow \frac{d}{dx}\frac{y'}{y} = \frac{y'}{y}$$
$$ \Longleftrightarrow \frac{dz}{dx} = z$$
where $z = y'/y$. The general solution is $ z = Ce^x$. Applying the initial condition gives $z(0) = y'(0)/y(0) = 1 \Rightarrow C = 1$. Now solve for $y$:
$$ \frac{y'}{y} = e^x $$
$$ \Rightarrow \int \frac{y'}{y} dx = \int e^x dx$$
$$ \Rightarrow \ln y = e^x + C_1$$
$$ \Rightarrow y = C_2 \exp(e^x) $$
Applying the initial condition gives $C_2 = 1/e$, i.e.,
$$ y = f(x) = \frac{\exp(e^x)}{e}. \quad (1)$$
Now verify the following statements:


*

*$f(x) = e^x$ has two solutions: wrong

*$f(x) = e^x$ has one solution: correct


Both statements can be seen from the following:
$$ (1) \Rightarrow e^{(e^x)} = ee^x$$
Taking $\ln$ both sides gives: $e^x = 1 + x$. Solving this equation is equivalent to finding all intersections between two curves $g(x) = e^x$ and $h(x) = x+1$. Note that


*

*the two curves $g(x)$ and $h(x)$ have one obvious intersection at $x = 0$,

*the $h(x)$ line is tangential to $g(x)$ at the intersection $x = 0$,

*going leftwards from $x = 0$ to $-\infty$: $g(x)$ decreases more slowly than $h(x)$ because the slope $g'(x) > h'(x)$ for all $x < 0$.
Thus, we conclude that there is one and only one intersection at $x = 0$, hence $f(x) = e^x$ has only one solution.


*$\lim_{x\to0} \frac{f(x)-1}{x} = \lim_{x\to0} \frac{f(x)-f(0)}{x - 0} = f'(0) = 1$: correct

*$\lim_{x\to-\infty} f(x) = \frac{1}{e} $: correct.
