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Ecuación diferencial 4*tan(x)*sec(x)^2*y'+2*sec(x)^2*y''+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                                      
         2     d               2    d                  
1 + 2*sec (x)*---(y(x)) + 4*sec (x)*--(y(x))*tan(x) = 0
                2                   dx                 
              dx                                       
$$4 \tan{\left(x \right)} \sec^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + 2 \sec^{2}{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} + 1 = 0$$
4*tan(x)*sec(x)^2*y' + 2*sec(x)^2*y'' + 1 = 0
Respuesta [src]
               2       2    2       2    2                                                                      
            sin (x)   x *cos (x)   x *sin (x)           2              2                         x*cos(x)*sin(x)
y(x) = C1 + ------- - ---------- - ---------- + C2*x*cos (x) + C2*x*sin (x) + C2*cos(x)*sin(x) - ---------------
               8          8            8                                                                4       
$$y{\left(x \right)} = C_{1} + C_{2} x \sin^{2}{\left(x \right)} + C_{2} x \cos^{2}{\left(x \right)} + C_{2} \sin{\left(x \right)} \cos{\left(x \right)} - \frac{x^{2} \sin^{2}{\left(x \right)}}{8} - \frac{x^{2} \cos^{2}{\left(x \right)}}{8} - \frac{x \sin{\left(x \right)} \cos{\left(x \right)}}{4} + \frac{\sin^{2}{\left(x \right)}}{8}$$
Clasificación
nth order reducible