y^{''}+2y^'+y=0,y(1)=6

解答

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6

y = (6 e - c_{2}) e^{- t} + c_{2} t e^{- t}

求解步骤

y^{''} (t) + 2 y^{'} (t) + y = 0

求解线性方程:

y = (6 e - c_{2}) e^{- t} + c_{2} t e^{- t}

y = (6 e - c_{2}) e^{- t} + c_{2} t e^{- t}

y^{^{''''}} + 7 y^{^{'''}} + 14 y^{^{''}} - 22 y^{^{'}} - 84 y = 0

(6 d^{3} + 5 d^{2}) y = 0

\frac{d ^{4} y}{d t ^{4}} + 2 y^{^{''}} + y = 0

y^{^{''}} = 4 y, y (0) = 1, y (1) = 0

y^{^{''}} + 36 y = 0, y (0) = cos (6 x)