2y^{''}+14y^'+12y=0

解答

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0

y = c_{1} e^{- t} + c_{2} e^{- 6 t}

求解步骤

2 y^{''} (t) + 14 y^{'} (t) + 12 y = 0

求解线性方程:

y = c_{1} e^{- t} + c_{2} e^{- 6 t}

y = c_{1} e^{- t} + c_{2} e^{- 6 t}

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

y^{^{''}} - 2 y^{^{'}} = - 4 y

7 y^{^{''}} - 8 y^{^{'}} + y = 0