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Ecuación diferencial dx*(-20*x*y^2+6*x)+dy*(-20*x^2*y+3*y^2)=0

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v

Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

Ha introducido [src]
            2         2    d              2 d                
6*x - 20*x*y (x) + 3*y (x)*--(y(x)) - 20*x *--(y(x))*y(x) = 0
                           dx               dx               
$$- 20 x^{2} y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} - 20 x y^{2}{\left(x \right)} + 6 x + 3 y^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
-20*x^2*y*y' - 20*x*y^2 + 6*x + 3*y^2*y' = 0
Respuesta [src]
                               /               /20    2 \\        
                               |            12*|-- - ---||        
                               |               |3      2||        
             2 /20    2 \    4 |800    80      \     C1 /|        
            x *|-- - ---|   x *|--- - --- + -------------|        
               |3      2|      | 3      2          2     |        
               \     C1 /      \      C1         C1      /    / 6\
y(x) = C1 + ------------- + ------------------------------ + O\x /
                  2                     24*C1                     
$$y{\left(x \right)} = \frac{x^{4} \left(\frac{800}{3} + \frac{12 \left(\frac{20}{3} - \frac{2}{C_{1}^{2}}\right)}{C_{1}^{2}} - \frac{80}{C_{1}^{2}}\right)}{24 C_{1}} + \frac{x^{2} \left(\frac{20}{3} - \frac{2}{C_{1}^{2}}\right)}{2} + C_{1} + O\left(x^{6}\right)$$
Gráfico para el problema de Cauchy
Clasificación
factorable
1st exact
1st power series
lie group
1st exact Integral
Respuesta numérica [src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 0.856896369722633)
(-5.555555555555555, 1.0738462933708053)
(-3.333333333333333, 1.6427429629153116)
(-1.1111111111111107, 7.157207331464997)
(1.1111111111111107, 8.100035731179602)
(3.333333333333334, 3.1933833808213433e-248)
(5.555555555555557, 3.1237768967464496e-33)
(7.777777777777779, 8.388243567719172e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)