1/8 y^{''}+4y^'+12y=0,y(0)=0,y^'(0)=4

解答

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 4

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}

求解步骤

\frac{1}{8} y^{''} (t) + 4 y^{'} (t) + 12 y = 0

求解线性方程:

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}

y = \frac{1}{2 10} e^{4 (10 - 4) t} - \frac{1}{2 10} e^{- 4 (4 + 10) t}