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Ecuación diferencial y''''+5*y''+4*y=sin(x)*cos(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:

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Solución

Ha introducido [src]
             2           4                        
            d           d                         
4*y(x) + 5*---(y(x)) + ---(y(x)) = cos(2*x)*sin(x)
             2           4                        
           dx          dx                         
$$4 y{\left(x \right)} + 5 \frac{d^{2}}{d x^{2}} y{\left(x \right)} + \frac{d^{4}}{d x^{4}} y{\left(x \right)} = \sin{\left(x \right)} \cos{\left(2 x \right)}$$
4*y + 5*y'' + y'''' = sin(x)*cos(2*x)
Respuesta [src]
                                                                                            /                   3                   2\       
         sin(7*x)   sin(5*x)   13*sin(3*x)                               /     x \          |     (1 - cos(2*x))    4*(1 - cos(2*x)) |       
y(x) = - -------- + -------- + ----------- + C3*sin(2*x) + C4*cos(2*x) + |C1 + --|*cos(x) + |C2 - --------------- + -----------------|*sin(x)
           240        144          360                                   \     12/          \            30                 45       /       
$$y{\left(x \right)} = C_{3} \sin{\left(2 x \right)} + C_{4} \cos{\left(2 x \right)} + \left(C_{1} + \frac{x}{12}\right) \cos{\left(x \right)} + \left(C_{2} - \frac{\left(1 - \cos{\left(2 x \right)}\right)^{3}}{30} + \frac{4 \left(1 - \cos{\left(2 x \right)}\right)^{2}}{45}\right) \sin{\left(x \right)} + \frac{13 \sin{\left(3 x \right)}}{360} + \frac{\sin{\left(5 x \right)}}{144} - \frac{\sin{\left(7 x \right)}}{240}$$
Clasificación
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral