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Ecuaciones diferenciales/ y=e^((x+1)/x),y*(1-ln(y))*y"+1+ln(y)*(y')^2=0

Ecuación diferencial y=e^((x+1)/x),y*(1-ln(y))*y"+1+ln(y)*(y')^2=0

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v

Para el problema de Cauchy:

y() =
y'() =
y''() =
y'''() =
y''''() =

Gráfico:

interior superior
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Solución

Ha introducido [src] 
         1 + x                                                             
         -----                2                               2            
           x        /d       \                               d             
y(x) = (e     , 1 + |--(y(x))| *log(y(x)) + (1 - log(y(x)))*---(y(x))*y(x))
                    \dx      /                                2            
                                                            dx
$$y{\left(x \right)} = \left( e^{\frac{x + 1}{x}}, \ \left(1 - \log{\left(y{\left(x \right)} \right)}\right) y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \log{\left(y{\left(x \right)} \right)} \left(\frac{d}{d x} y{\left(x \right)}\right)^{2} + 1\right)$$ 
y = (exp((x + 1)/x, (1 - log(y))*y*y'' + log(y)*y'^2 + 1))
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