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Ecuación diferencial y''-2y'+5y=e^x(x-1)+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                               
    d                    d                    x           
- 2*--(y(x)) + 5*y(x) + ---(y(x)) = (-1 + x)*e  + cos(2*x)
    dx                    2                               
                        dx                                
$$5 y{\left(x \right)} - 2 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(x - 1\right) e^{x} + \cos{\left(2 x \right)}$$
5*y - 2*y' + y'' = (x - 1)*exp(x) + cos(2*x)
Respuesta [src]
         4*sin(2*x)   cos(2*x)   /  1   x                            \  x
y(x) = - ---------- + -------- + |- - + - + C1*sin(2*x) + C2*cos(2*x)|*e 
             17          17      \  4   4                            /   
$$y{\left(x \right)} = \left(C_{1} \sin{\left(2 x \right)} + C_{2} \cos{\left(2 x \right)} + \frac{x}{4} - \frac{1}{4}\right) e^{x} - \frac{4 \sin{\left(2 x \right)}}{17} + \frac{\cos{\left(2 x \right)}}{17}$$
Clasificación
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral