y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2

解答

y^{^{''}} + 2 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}

求解步骤

y^{''} (t) + 2 y^{'} (t) - 24 y = 0

求解线性方程:

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}