How to solve the ODE $ f(x)f''(x)-f(x)f'(x)-{f'(x)}^2=0$? 
Solve the differential equation $$f(x)f''(x)-f(x)f'(x)-f'(x)^2=0$$ with $f(0) = 0 = f'(x)$.


my attempt:-
Let $f(x)f'(x)=z$
so we have $f'(x)^2+f(x)f''(x)= \frac{dz}{dx}$, we get
substituting in :- $ f(x)f''(x)-f(x)f'(x)-{f'(x)}^2=0$,we get
$$f(x)f''(x)-z-\left(\frac{dz}{dx}-f(x)f''(x)\right)=0 ,$$
beyond which I'm stuck. Could someone help, please?
 A: $$f(x)f''(x)-f(x)f'(x)-{f'(x)}^2=0$$
$$\dfrac {f''(x)}{f'(x)}-1=\dfrac {f'(x)}{f(x)}$$
Integrate to reduce the order:
$$\ln (f'(x))-x=\ln f(x)+C$$
$$\ln \left (\dfrac {f'(x)}{f(x)}\right)=x+C$$
$$(\ln  f(x))'=ce^x$$
Integrate again.
A: If a non-constant solution exists and can be found symbolically, then the substitution $f'(x)=u(f(x))$ with $f''(x)=u'(f(x))u(f(x))$ often helps sort out the terms. Here it leads to
$$
fu(f)u'(f)-fu(f)-u(f)^2=0
\\
u'(f)=1+\frac{u(f)}{f}.
$$
This now is a (degree-)homogeneous ODE, with the substitution $u(f)=fv(f)$ giving
$$
fv'(f)=1\implies v(f)=a+\ln|f|, ~~ f'(x)=u(f)=af+f\ln|f|
$$
which now can be solved by setting $g=\ln|f|$, $g'(x)=a+g(x)$, ...
A: Inspection shows that constant functions are solutions and in particular that $f(x) = 0$ solves the initial value problem, but we can find the general solution to the o.d.e.
Hint That each term has the same number of factors $f^{(k)}(x)$ suggests substituting $$f(x) = \exp u(x) ,$$ which transforms the equation to the second-order, constant-coefficient linear o.d.e. $$u''(x) - u'(x) = 0 ,$$ which has solution $u(x) = A e^x + B$. Substituting gives
$$f(x) = e^{A e^x + B} = C e^{A e^x},$$ and allowing $C = 0$ recovers the zero solution.
