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Ecuación diferencial y'''+4y''-12y'=8e^(2x)(cosx)(senx)

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                    2*x       
- 12*--(y(x)) + 4*---(y(x)) + ---(y(x)) = 8*cos(x)*e   *sin(x)
     dx             2           3                             
                  dx          dx                              
$$- 12 \frac{d}{d x} y{\left(x \right)} + 4 \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \frac{d^{3}}{d x^{3}} y{\left(x \right)} = 8 e^{2 x} \sin{\left(x \right)} \cos{\left(x \right)}$$
-12*y' + 4*y'' + y''' = 8*exp(2*x)*sin(x)*cos(x)
Respuesta [src]
                -6*x   /     5*sin(2*x)   3*cos(2*x)\  2*x
y(x) = C1 + C3*e     + |C2 - ---------- - ----------|*e   
                       \         68           68    /     
$$y{\left(x \right)} = C_{1} + C_{3} e^{- 6 x} + \left(C_{2} - \frac{5 \sin{\left(2 x \right)}}{68} - \frac{3 \cos{\left(2 x \right)}}{68}\right) e^{2 x}$$
Clasificación
nth linear constant coeff variation of parameters
nth order reducible
nth linear constant coeff variation of parameters Integral