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Ecuación diferencial y''+2y'+5y=x^2-5x+2+e^(x)

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Para el problema de Cauchy:

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

Gráfico:

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Solución

Ha introducido [src]
                        2                          
  d                    d               2          x
2*--(y(x)) + 5*y(x) + ---(y(x)) = 2 + x  - 5*x + e 
  dx                    2                          
                      dx                           
$$5 y{\left(x \right)} + 2 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = x^{2} - 5 x + e^{x} + 2$$
5*y + 2*y' + y'' = x^2 - 5*x + exp(x) + 2
Respuesta [src]
                     2    x                                  
        98   29*x   x    e                                 -x
y(x) = --- - ---- + -- + -- + (C1*sin(2*x) + C2*cos(2*x))*e  
       125    25    5    8                                   
$$y{\left(x \right)} = \frac{x^{2}}{5} - \frac{29 x}{25} + \left(C_{1} \sin{\left(2 x \right)} + C_{2} \cos{\left(2 x \right)}\right) e^{- x} + \frac{e^{x}}{8} + \frac{98}{125}$$
Clasificación
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