2y^{''}+16y^'+8y=0,y(0)=5,y^'(0)=0

解

2 y^{^{''}} + 16 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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

解答ステップ

2 y^{''} (t) + 16 y^{'} (t) + 8 y = 0

線形 ODE を解く:

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

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