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Ecuación diferencial y'''+2y''+5y'-26y=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                        3          
              d            d           d           
-26*y(x) + 2*---(y(x)) + 5*--(y(x)) + ---(y(x)) = 0
               2           dx           3          
             dx                       dx           
$$- 26 y{\left(x \right)} + 5 \frac{d}{d x} y{\left(x \right)} + 2 \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \frac{d^{3}}{d x^{3}} y{\left(x \right)} = 0$$
-26*y + 5*y' + 2*y'' + y''' = 0
Respuesta [src]
           2*x                                -2*x
y(x) = C3*e    + (C1*sin(3*x) + C2*cos(3*x))*e    
$$y{\left(x \right)} = C_{3} e^{2 x} + \left(C_{1} \sin{\left(3 x \right)} + C_{2} \cos{\left(3 x \right)}\right) e^{- 2 x}$$
Clasificación
nth linear constant coeff homogeneous