Sr Examen

Ecuación diferencial y"-tg(x)y'+5y=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]
                             2          
         d                  d           
5*y(x) - --(y(x))*tan(x) + ---(y(x)) = 0
         dx                  2          
                           dx           
$$5 y{\left(x \right)} - \tan{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0$$
5*y - tan(x)*y' + y'' = 0
Respuesta [src]
          /       2       4      3             4    2   \        /       2                 3           2    2       3    3   \        
          |    5*x    25*x    5*x *tan(x)   5*x *tan (x)|        |    5*x    x*tan(x)   5*x *tan(x)   x *tan (x)   x *tan (x)|    / 6\
y(x) = C2*|1 - ---- + ----- - ----------- - ------------| + C1*x*|1 - ---- + -------- - ----------- + ---------- + ----------| + O\x /
          \     2       24         6             24     /        \     6        2            12           6            24    /        
$$y{\left(x \right)} = C_{2} \left(- \frac{5 x^{4} \tan^{2}{\left(x \right)}}{24} + \frac{25 x^{4}}{24} - \frac{5 x^{3} \tan{\left(x \right)}}{6} - \frac{5 x^{2}}{2} + 1\right) + C_{1} x \left(\frac{x^{3} \tan^{3}{\left(x \right)}}{24} - \frac{5 x^{3} \tan{\left(x \right)}}{12} + \frac{x^{2} \tan^{2}{\left(x \right)}}{6} - \frac{5 x^{2}}{6} + \frac{x \tan{\left(x \right)}}{2} + 1\right) + O\left(x^{6}\right)$$
Clasificación
2nd power series ordinary