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Ecuación diferencial y'''-2y''-4y'+8y=x^2exp(-2x)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Para el problema de Cauchy:

y() =
y'() =
y''() =
y'''() =
y''''() =

Gráfico:

interior superior

Solución

Ha introducido [src]
                   2                    3                 
    d             d                    d           2  -2*x
- 4*--(y(x)) - 2*---(y(x)) + 8*y(x) + ---(y(x)) = x *e    
    dx             2                    3                 
                 dx                   dx                  
$$8 y{\left(x \right)} - 4 \frac{d}{d x} y{\left(x \right)} - 2 \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \frac{d^{3}}{d x^{3}} y{\left(x \right)} = x^{2} e^{- 2 x}$$
8*y - 4*y' - 2*y'' + y''' = x^2*exp(-2*x)
Respuesta [src]
                          /      2    3      \      
                    2*x   |     x    x    3*x|  -2*x
y(x) = (C1 + C2*x)*e    + |C3 + -- + -- + ---|*e    
                          \     32   48   128/      
$$y{\left(x \right)} = \left(C_{1} + C_{2} x\right) e^{2 x} + \left(C_{3} + \frac{x^{3}}{48} + \frac{x^{2}}{32} + \frac{3 x}{128}\right) e^{- 2 x}$$
Clasificación
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral