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

解答

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

y = (2 + \frac{2 - 3}{3}) e^{(- 1 + 3) t} - \frac{2 - 3}{3} e^{(- 1 - 3) t}

求解步骤

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

求解线性方程:

y = (2 + \frac{2 - 3}{3}) e^{(- 1 + 3) t} - \frac{2 - 3}{3} e^{(- 1 - 3) t}

y = (2 + \frac{2 - 3}{3}) e^{(- 1 + 3) t} - \frac{2 - 3}{3} e^{(- 1 - 3) t}

9 y^{^{''}} + 12 y^{^{'}} + 29 y = 0

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0