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

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v

Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

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