y^{'''}+2y^{''}+y^'=2-24e^x+40e^{5x}

解答

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5 x}

y = c_{1} + c_{2} e^{- x} + c_{3} x e^{- x} + 2 x - 6 e^{x} + \frac{2}{9} e^{5 x}

求解步骤

y^{'''} (x) + 2 y^{''} (x) + y^{'} (x) = 2 - 24 e^{x} + 40 e^{5 x}

求解线性方程:

y = c_{1} + c_{2} e^{- x} + c_{3} x e^{- x} + 2 x - 6 e^{x} + \frac{2}{9} e^{5 x}

y = c_{1} + c_{2} e^{- x} + c_{3} x e^{- x} + 2 x - 6 e^{x} + \frac{2}{9} e^{5 x}

y^{^{''}} + 2 y = cos (2) t + sin (2) t

y^{^{''}} + 25 y = 7 sec (5 x)

2 y^{^{''}} + 3 y^{^{'}} + y = x^{2}, y (0) = 0, y^{^{'}} (0) = 1

(d^{2} - 1) y = sin^{2} (x)

41 H (x) - 8 u^{^{'}} + u^{^{''}} = 0