(D^3-D^2-D+1)y=0

解答

(D^{3} - D^{2} - D + 1) y = 0

y = c_{1} e^{- x} + c_{2} e^{x} + c_{3} x e^{x}

求解步骤

(D^{3} - D^{2} - D + 1) y = 0

求解线性方程:

y = c_{1} e^{- x} + c_{2} e^{x} + c_{3} x e^{x}

y = c_{1} e^{- x} + c_{2} e^{x} + c_{3} x e^{x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y^{0} = 0, y^{^{'}} 0 = 1

4 \frac{d ^{2} y}{d x ^{2}} + 32 \frac{d y}{d x} + 64 y = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 8, y^{^{'}} (0) = 9

y^{^{''}} + 6 y = 0, y (0) = 0, y (1) = 0

y^{^{''}} + 4 y^{^{'}} - 32 y = 0, y (0) = a, y^{^{'}} (0) = - 40