x^{''}+8x^'+11x=0,x(0)=0,x^'(0)=3

解答

x^{^{''}} + 8 x^{^{'}} + 11 x = 0, x (0) = 0, x^{^{'}} (0) = 3

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

求解步骤

x^{''} (t) + 8 x^{'} (t) + 11 x = 0

求解线性方程:

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

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

y^{^{''}} - 8 y^{^{''}} - 9 y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = \frac{1}{4}

y^{^{''}} + (3 - a) y^{^{'}} + 2 (a - 1) y = 0

\frac{d ^{2} y}{d x ^{2}} - 12 \frac{d y}{d x} - 64 y = 0

\frac{10}{32} y^{^{''}} + \frac{1}{6} y = 0, y (0) = \frac{1}{6}, y^{^{'}} (0) = - 1