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Ecuación diferencial y'''-5y''+8y'-4y=(2x-5)exp(x)

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:

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Solución

Ha introducido [src]
      2                                 3                      
     d                     d           d                      x
- 5*---(y(x)) - 4*y(x) + 8*--(y(x)) + ---(y(x)) = (-5 + 2*x)*e 
      2                    dx           3                      
    dx                                dx                       
$$- 4 y{\left(x \right)} + 8 \frac{d}{d x} y{\left(x \right)} - 5 \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \frac{d^{3}}{d x^{3}} y{\left(x \right)} = \left(2 x - 5\right) e^{x}$$
-4*y + 8*y' - 5*y'' + y''' = (2*x - 5)*exp(x)
Respuesta [src]
       /      2                    x\  x
y(x) = \C1 + x  - x + (C2 + C3*x)*e /*e 
$$y{\left(x \right)} = \left(C_{1} + x^{2} - x + \left(C_{2} + C_{3} x\right) e^{x}\right) e^{x}$$
Clasificación
nth linear constant coeff undetermined coefficients
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