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Ecuación diferencial dx*(x+sin(2*x)/y)+dy*(y-sin(x)^2/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    d           
                               sin (x)*--(y(x))    
    sin(2*x)   d                       dx          
x + -------- + --(y(x))*y(x) - ---------------- = 0
      y(x)     dx                    2             
                                    y (x)          
$$x + y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + \frac{\sin{\left(2 x \right)}}{y{\left(x \right)}} - \frac{\sin^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)}}{y^{2}{\left(x \right)}} = 0$$
x + y*y' + sin(2*x)/y - sin(x)^2*y'/y^2 = 0
Respuesta [src]
                              /                  /    4 \ /     2 \\        
                              |                3*|1 + --|*|-1 - --||        
                            4 |     12    6      \    C1/ \     C1/|        
             2 /     2 \   x *|8 - --- - --- + --------------------|        
            x *|-1 - --|      |      3     2            C1         |        
               \     C1/      \    C1    C1                        /    / 6\
y(x) = C1 + ------------ + ----------------------------------------- + O\x /
                2*C1                              2                         
                                             24*C1                          
$$y{\left(x \right)} = \frac{x^{4} \left(8 + \frac{3 \left(-1 - \frac{2}{C_{1}}\right) \left(1 + \frac{4}{C_{1}}\right)}{C_{1}} - \frac{6}{C_{1}^{2}} - \frac{12}{C_{1}^{3}}\right)}{24 C_{1}^{2}} + \frac{x^{2} \left(-1 - \frac{2}{C_{1}}\right)}{2 C_{1}} + C_{1} + O\left(x^{6}\right)$$
Clasificación
factorable
1st exact
1st power series
lie group
1st exact Integral