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Ecuación diferencial sqrt(y'+1)(yy''-(y')^2)=yy'sqrt(yy')

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Para el problema de Cauchy:

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

Gráfico:

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Solución

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