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Ecuación diferencial a^2*y''+a*y'-y=-2*a^2*cos(x)+a

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v

Para el problema de Cauchy:

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

Gráfico:

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Solución

Ha introducido [src]
                          2                        
          d           2  d                 2       
-y(x) + a*--(y(x)) + a *---(y(x)) = a - 2*a *cos(x)
          dx              2                        
                        dx                         
$$a^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)} + a \frac{d}{d x} y{\left(x \right)} - y{\left(x \right)} = - 2 a^{2} \cos{\left(x \right)} + a$$
a^2*y'' + a*y' - y = -2*a^2*cos(x) + a
Respuesta [src]
                  /       ___\         /       ___\                                                
                x*\-1 + \/ 5 /       x*\-1 - \/ 5 /                                                
                --------------       --------------       3               2               4        
                     2*a                  2*a          2*a *sin(x)     2*a *cos(x)     2*a *cos(x) 
y(x) = -a + C1*e               + C2*e               - ------------- + ------------- + -------------
                                                           4      2        4      2        4      2
                                                      1 + a  + 3*a    1 + a  + 3*a    1 + a  + 3*a 
$$y{\left(x \right)} = C_{1} e^{\frac{x \left(-1 + \sqrt{5}\right)}{2 a}} + C_{2} e^{\frac{x \left(- \sqrt{5} - 1\right)}{2 a}} + \frac{2 a^{4} \cos{\left(x \right)}}{a^{4} + 3 a^{2} + 1} - \frac{2 a^{3} \sin{\left(x \right)}}{a^{4} + 3 a^{2} + 1} + \frac{2 a^{2} \cos{\left(x \right)}}{a^{4} + 3 a^{2} + 1} - a$$
Clasificación
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral