Solve differential equation $f''''(x)=f'''(x)f''(x)f'(x)f(x)$ I met this DE recently, and I am utterly befuddled at how to solve it
$$f''''(x)=f'''(x)f''(x)f'(x)f(x)$$
I tried this:
$$\frac{f''''(x)}{f'''(x)}=f''(x)f'(x)f(x)$$
$$\ln|f'''(x)|=c_1+\int f(x)f'(x)f''(x)dx$$
I do not know how to solve the right side, though. Integration by parts?
Plaese help.
 A: Here are the first few terms of the series solution at $0$, according to Maple.
$$
f \left( x \right) ={\it a_0}+{\it a_1}\,x+{\it a_2}\,{x}^{2}+{
\it a_3}\,{x}^{3}\\+{\frac {{\it a_3}\,{\it a_2}\,{\it a_1}\,{\it 
a_0}}{2}}\,{x}^{4}\\+ \left( {\frac {{{\it a_0}}^{2}{{\it a_1}}^{2}{{
\it a_2}}^{2}{\it a_3}}{5}}+{\frac {3\,{\it a_0}\,{\it a_1}\,{{
\it a_3}}^{2}}{10}}+{\frac {{\it a_0}\,{{\it a_2}}^{2}{\it a_3}}{5
}}+{\frac {{{\it a_1}}^{2}{\it a_2}\,{\it a_3}}{10}} \right) {x}^{5
}\\+ \left( {\frac {{{\it a_0}}^{3}{{\it a_1}}^{3}{{\it a_2}}^{3}{
\it a_3}}{15}}+{\frac {2\,{{\it a_3}}^{2}{\it a_2}\,{{\it a_1}}^{2
}{{\it a_0}}^{2}}{5}}+{\frac {{{\it a_0}}^{2}{\it a_1}\,{{\it a_2}
}^{3}{\it a_3}}{5}}+{\frac {{\it a_0}\,{{\it a_1}}^{3}{{\it a_2}}^
{2}{\it a_3}}{10}}+{\frac {3\,{{\it a_3}}^{2}{\it a_0}\,{\it a_2}
}{10}}+{\frac {{{\it a_1}}^{2}{{\it a_3}}^{2}}{10}}+{\frac {{{\it 
a_2}}^{2}{\it a_1}\,{\it a_3}}{10}} \right) {x}^{6}\\+\dots 
$$
With $4$ arbitrary constants, as expected.  But, again as expected, it is not a linear combination of 4 independent solutions.
added  How'd you get this?
Maple.  Outputs are long, so I omit them.
>DE:=diff(f(x),x,x,x,x)=diff(f(x),x,x,x)*diff(f(x),x,x)*diff(f(x),x)*f(x);
>Order:=7;SS:=dsolve({DE,f(0)=a_0,D(f)(0)=a_1,
(D@@2)(f)(0)=2*a_2,(D@@3)(f)(0)=6*a_3},f(x),series,x=0);
>latex(%);
A: Hint:
Let $u=\dfrac{df}{dx}$ ,
Then $\dfrac{d^2f}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{df}\dfrac{df}{dx}=u\dfrac{du}{df}$
$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(u\dfrac{du}{df}\right)=\dfrac{d}{df}\left(u\dfrac{du}{df}\right)\dfrac{df}{dx}=\left(u\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^2\right)u=u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2$
$\dfrac{d^4f}{dx^4}=\dfrac{d}{dx}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)=\dfrac{d}{df}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)\dfrac{df}{dx}=\left(u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3\right)u=u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3$
$\therefore u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3=\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)u\dfrac{du}{df}uf$
$u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3=fu^3\dfrac{du}{df}\dfrac{d^2u}{df^2}+fu^2\left(\dfrac{du}{df}\right)^3$
$u^2\dfrac{d^3u}{df^3}+(4-fu^2)u\dfrac{du}{df}\dfrac{d^2u}{df^2}+(1-fu^2)\left(\dfrac{du}{df}\right)^3=0$
