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Ecuación diferencial y"+4*y'+5*y=(e^(-2x))*(1+sec(x)^2)

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Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

Ha introducido [src]
                        2                            
  d                    d          /       2   \  -2*x
4*--(y(x)) + 5*y(x) + ---(y(x)) = \1 + sec (x)/*e    
  dx                    2                            
                      dx                             
$$5 y{\left(x \right)} + 4 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(\sec^{2}{\left(x \right)} + 1\right) e^{- 2 x}$$
5*y + 4*y' + y'' = (sec(x)^2 + 1)*exp(-2*x)
Respuesta [src]
       /            /     log(1 + sin(x))   log(-1 + sin(x))\       \  -2*x
y(x) = |C2*cos(x) + |C1 + --------------- - ----------------|*sin(x)|*e    
       \            \            2                 2        /       /      
$$y{\left(x \right)} = \left(C_{2} \cos{\left(x \right)} + \left(C_{1} - \frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{2}\right) \sin{\left(x \right)}\right) e^{- 2 x}$$
Clasificación
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral