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Ecuación diferencial y''-8y'+25y=3x^2+4x

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             2      
- 8*--(y(x)) + 25*y(x) + ---(y(x)) = 3*x  + 4*x
    dx                     2                   
                         dx                    
$$25 y{\left(x \right)} - 8 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 3 x^{2} + 4 x$$
25*y - 8*y' + y'' = 3*x^2 + 4*x
Respuesta [src]
                  2                                           
        1034   3*x    148*x                                4*x
y(x) = ----- + ---- + ----- + (C1*sin(3*x) + C2*cos(3*x))*e   
       15625    25     625                                    
$$y{\left(x \right)} = \frac{3 x^{2}}{25} + \frac{148 x}{625} + \left(C_{1} \sin{\left(3 x \right)} + C_{2} \cos{\left(3 x \right)}\right) e^{4 x} + \frac{1034}{15625}$$
Clasificación
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral