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Ecuación diferencial dx*(x*y^2+y)+dy*(x^2*y+x-1)=0

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]
  d             2        d           2 d                       
- --(y(x)) + x*y (x) + x*--(y(x)) + x *--(y(x))*y(x) + y(x) = 0
  dx                     dx            dx                      
$$x^{2} y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + x y^{2}{\left(x \right)} + x \frac{d}{d x} y{\left(x \right)} + y{\left(x \right)} - \frac{d}{d x} y{\left(x \right)} = 0$$
x^2*y*y' + x*y^2 + x*y' + y - y' = 0
Respuesta [src]
                   ______________________
                  /      2             2 
            1   \/  1 + x  - 2*x + C1*x  
       -1 + - - -------------------------
            x               x            
y(x) = ----------------------------------
                       x                 
$$y{\left(x \right)} = \frac{-1 - \frac{\sqrt{C_{1} x^{2} + x^{2} - 2 x + 1}}{x} + \frac{1}{x}}{x}$$
                   ______________________
                  /      2             2 
            1   \/  1 + x  - 2*x + C1*x  
       -1 + - + -------------------------
            x               x            
y(x) = ----------------------------------
                       x                 
$$y{\left(x \right)} = \frac{-1 + \frac{\sqrt{C_{1} x^{2} + x^{2} - 2 x + 1}}{x} + \frac{1}{x}}{x}$$
Clasificación
1st exact
1st power series
lie group
1st exact Integral