y^{''}+6y^'+8y=0,y(0)= 2/5 ,y^'(0)= 2/5

解答

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = \frac{2}{5}, y^{^{'}} (0) = \frac{2}{5}

y = (\frac{2}{5} + \frac{3}{5}) e^{- 2 t} - \frac{3}{5} e^{- 4 t}

求解步骤

y^{''} (t) + 6 y^{'} (t) + 8 y = 0

求解线性方程:

y = (\frac{2}{5} + \frac{3}{5}) e^{- 2 t} - \frac{3}{5} e^{- 4 t}

y = (\frac{2}{5} + \frac{3}{5}) e^{- 2 t} - \frac{3}{5} e^{- 4 t}

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y^{^{'}} (0) = 0, y (0) = 5

- 2 y^{^{''}} + 8 y^{^{'}} = 0, y (- 1) = 2, y^{^{'}} (2) = 2

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = 1, y^{^{'}} (0) = 9

y^{^{''}} = - y^{^{'}} - y

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1