Resolver ecuación diferencial$f''''(x)=f'''(x)f''(x)f'(x)f(x)$

Question

Me encontré con este DE recientemente, y estoy completamente aturdido en cómo resolverlo $$f''''(x)=f'''(x)f''(x)f'(x)f(x)$ $ Intenté esto: $$\frac{f''''(x)}{f'''(x)}=f''(x)f'(x)f(x)$ $ $$\ln|f'''(x)|=c_1+\int f(x)f'(x)f''(x)dx$ $ No sé cómo resolver el lado correcto, sin embargo . ¿Integración por partes?

Ayuda plaese

Answer 1

Aquí están los primeros términos de la serie de la solución en $0$, según Arce. $$ f \left( x \right) ={\it a_0}+{\it a_1}\,x+{\it a_2}\,{x}^{2}+{ \se a_3}\,{x}^{3}\\+{\frac {{\it a_3}\,{\it a_2}\,{\it a_1}\,{\it a_0}}{2}}\,{x}^{4}\\+ \left( {\frac {{{\it a_0}}^{2}{{\it a_1}}^{2}{{ \se a_2}}^{2}{\it a_3}}{5}}+{\frac {3\,{\it a_0}\,{\it a_1}\,{{ \se a_3}}^{2}}{10}}+{\frac {{\it a_0}\,{{\it a_2}}^{2}{\it a_3}}{5 }}+{\frac {{{\it a_1}}^{2}{\it a_2}\,{\it a_3}}{10}} \right) {x}^{5 }\\+ \left( {\frac {{{\it a_0}}^{3}{{\it a_1}}^{3}{{\it a_2}}^{3}{ \se a_3}}{15}}+{\frac {2\,{{\it a_3}}^{2}{\it a_2}\,{{\it a_1}}^{2 }{{\it a_0}}^{2}}{5}}+{\frac {{{\it a_0}}^{2}{\it a_1}\,{{\it a_2} }^{3}{\it a_3}}{5}}+{\frac {{\it a_0}\,{{\it a_1}}^{3}{{\it a_2}}^ {2}{\it a_3}}{10}}+{\frac {3\,{{\it a_3}}^{2}{\it a_0}\,{\it a_2} }{10}}+{\frac {{{\it a_1}}^{2}{{\it a_3}}^{2}}{10}}+{\frac {{{\it a_2}}^{2}{\it a_1}\,{\it a_3}}{10}} \right) {x}^{6}\\+\dots $$ Con $4$ constantes arbitrarias, como se esperaba. Pero, de nuevo, como era de esperar, no es una combinación lineal de 4 soluciones independientes.

agregó ¿Cómo se consigue esto?
De arce. Las salidas son largas, así que omito.

>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(%);

Answer 2

Sugerencia:

Que <span class="math-container">$u=\dfrac{df}{dx}$</span> ,

Entonces <span class="math-container">$\dfrac{d^2f}{dx^2}=\dfrac{du}{dx}=\dfrac{du}{df}\dfrac{df}{dx}=u\dfrac{du}{df}$</span>

<span class="math-container">$\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(u\dfrac{du}{df}\right)=\dfrac{d}{df}\left(u\dfrac{du}{df}\right)\dfrac{df}{dx}=\left(u\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^2\right)u=u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2$</span>

<span class="math-container">$\dfrac{d^4f}{dx^4}=\dfrac{d}{dx}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)=\dfrac{d}{df}\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)\dfrac{df}{dx}=\left(u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3\right)u=u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3$</span>

<span class="math-container">$\therefore u^3\dfrac{d^3u}{df^3}+4u^2\dfrac{du}{df}\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^3=\left(u^2\dfrac{d^2u}{df^2}+u\left(\dfrac{du}{df}\right)^2\right)u\dfrac{du}{df}uf$</span>

<span class="math-container">$u^2\dfrac{d^3u}{df^3}+4u\dfrac{du}{df}\dfrac{d^2u}{df^2}+\left(\dfrac{du}{df}\right)^3=fu^3\dfrac{du}{df}\dfrac{d^2u}{df^2}+fu^2\left(\dfrac{du}{df}\right)^3$</span>

<span class="math-container">$u^2\dfrac{d^3u}{df^3}+(4-fu^2)u\dfrac{du}{df}\dfrac{d^2u}{df^2}+(1-fu^2)\left(\dfrac{du}{df}\right)^3=0$</span>