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Ecuación diferencial dx*(x^3-3*x*y^2+2)-dy*(3*x^2*y-y^2)=0

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]
     3    2    d               2         2 d                
2 + x  + y (x)*--(y(x)) - 3*x*y (x) - 3*x *--(y(x))*y(x) = 0
               dx                          dx               
$$x^{3} - 3 x^{2} y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} - 3 x y^{2}{\left(x \right)} + y^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + 2 = 0$$
x^3 - 3*x^2*y*y' - 3*x*y^2 + y^2*y' + 2 = 0
Respuesta [src]
             2 /     8 \                  5 /     1408   6 \    4 /     640    3 \        
            x *|3 - ---|                 x *|27 - ---- - --|   x *|27 - ---- - --|        
               |      5|             3      |       10   C1|      |       10   C1|        
               \    C1 /   2*x   40*x       \     C1       /      \     C1       /    / 6\
y(x) = C1 + ------------ - --- - ----- + ------------------- + ------------------- + O\x /
                 2           2       8              4                 12*C1               
                           C1    3*C1           6*C1                                      
$$y{\left(x \right)} = - \frac{40 x^{3}}{3 C_{1}^{8}} + \frac{x^{5} \left(27 - \frac{6}{C_{1}} - \frac{1408}{C_{1}^{10}}\right)}{6 C_{1}^{4}} - \frac{2 x}{C_{1}^{2}} + \frac{x^{4} \left(27 - \frac{3}{C_{1}} - \frac{640}{C_{1}^{10}}\right)}{12 C_{1}} + \frac{x^{2} \left(3 - \frac{8}{C_{1}^{5}}\right)}{2} + C_{1} + O\left(x^{6}\right)$$
Clasificación
factorable
1st exact
1st power series
lie group
1st exact Integral