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Ecuación diferencial tgyy''=(y')^2

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

Ha introducido [src]
          2                                             2                          2                                                2
/d       \  /         2/ 2   \\   /         2/ 2   \\  d                 /d       \   2    /       2/ 2   \\    / 2   \   /d       \ 
|--(y(x))| *\2 + 2*tan \y (x)// + \2 + 2*tan \y (x)//*---(y(x))*y(x) + 8*|--(y(x))| *y (x)*\1 + tan \y (x)//*tan\y (x)/ = |--(y(x))| 
\dx      /                                              2                \dx      /                                       \dx      / 
                                                      dx                                                                             
$$8 \left(\tan^{2}{\left(y^{2}{\left(x \right)} \right)} + 1\right) y^{2}{\left(x \right)} \tan{\left(y^{2}{\left(x \right)} \right)} \left(\frac{d}{d x} y{\left(x \right)}\right)^{2} + \left(2 \tan^{2}{\left(y^{2}{\left(x \right)} \right)} + 2\right) y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \left(2 \tan^{2}{\left(y^{2}{\left(x \right)} \right)} + 2\right) \left(\frac{d}{d x} y{\left(x \right)}\right)^{2} = \left(\frac{d}{d x} y{\left(x \right)}\right)^{2}$$
8*(tan(y^2)^2 + 1)*y^2*tan(y^2)*y'^2 + (2*tan(y^2)^2 + 2)*y*y'' + (2*tan(y^2)^2 + 2)*y'^2 = y'^2
Clasificación
factorable