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Ecuación diferencial dx*(2*x*y^2+y+cos(x))+dy*(2^y*log(2)+2*x^2*y+x)=0

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

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