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Ecuación diferencial (x/(x^2+y^2)+(2xy)/(x^2+y^2)^2)dx+(1+y/(x^2+y^2)+(y^2-x^2)/(x^2+y^2)^2)dy

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                                                       
    x            dy*y(x)                    dy*y (x)                            dy*x                          2*x*y(x)          d           
---------- + ---------------- + -------------------------------- - -------------------------------- + ----------------------- + --(y(x)) = 0
 2    2          2       2          4       4            2  2          4       4            2  2       4    4         2  2      dx          
x  + y (x)   dx*x  + dx*y (x)   dx*x  + dx*y (x) + 2*dx*x *y (x)   dx*x  + dx*y (x) + 2*dx*x *y (x)   x  + y (x) + 2*x *y (x)               
$$- \frac{dy x^{2}}{dx x^{4} + 2 dx x^{2} y^{2}{\left(x \right)} + dx y^{4}{\left(x \right)}} + \frac{dy y^{2}{\left(x \right)}}{dx x^{4} + 2 dx x^{2} y^{2}{\left(x \right)} + dx y^{4}{\left(x \right)}} + \frac{dy y{\left(x \right)}}{dx x^{2} + dx y^{2}{\left(x \right)}} + \frac{2 x y{\left(x \right)}}{x^{4} + 2 x^{2} y^{2}{\left(x \right)} + y^{4}{\left(x \right)}} + \frac{x}{x^{2} + y^{2}{\left(x \right)}} + \frac{d}{d x} y{\left(x \right)} = 0$$
-dy*x^2/(dx*x^4 + 2*dx*x^2*y^2 + dx*y^4) + dy*y^2/(dx*x^4 + 2*dx*x^2*y^2 + dx*y^4) + dy*y/(dx*x^2 + dx*y^2) + 2*x*y/(x^4 + 2*x^2*y^2 + y^4) + x/(x^2 + y^2) + y' = 0
Clasificación
1st power series
lie group