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Ecuación diferencial (2x+y)dy=ydx+4lnydy

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                   d            d                        
--(y(x))*y(x) + 2*x*--(y(x)) = 4*--(y(x))*log(y(x)) + y(x)
dx                  dx           dx                       
$$2 x \frac{d}{d x} y{\left(x \right)} + y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = y{\left(x \right)} + 4 \log{\left(y{\left(x \right)} \right)} \frac{d}{d x} y{\left(x \right)}$$
2*x*y' + y*y' = y + 4*log(y)*y'
Respuesta [src]
1 - x - y(x) + 2*log(y(x))     
-------------------------- = C1
           2                   
          y (x)                
$$\frac{- x - y{\left(x \right)} + 2 \log{\left(y{\left(x \right)} \right)} + 1}{y^{2}{\left(x \right)}} = C_{1}$$
Clasificación
1st exact
1st power series
lie group
1st exact Integral