Ecuación diferencial ydx+(2xy-e^(-2y))dy=0

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Ha introducido [src]

  d         -2*y(x)       d                       
- --(y(x))*e        + 2*x*--(y(x))*y(x) + y(x) = 0
  dx                      dx

$$2 x y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + y{\left(x \right)} - e^{- 2 y{\left(x \right)}} \frac{d}{d x} y{\left(x \right)} = 0$$

2*x*y*y' + y - exp(-2*y)*y' = 0

Respuesta [src]

                2*y(x)     
-log(y(x)) + x*e       = C1

$$x e^{2 y{\left(x \right)}} - \log{\left(y{\left(x \right)} \right)} = C_{1}$$

Clasificación

1st exact

1st power series

lie group

1st exact Integral