8y^{''}+cy^'+128y=0,y(0)=5,y^'(0)=8

解答

8 y^{^{''}} + c y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

y = e^{\frac{t ( - c + c ^{2} - 4096 )}{16}} (5 + \frac{128 + 5 c - 5 c ^{2} - 4096}{2 c ^{2} - 4096}) - \frac{e ^{\frac{t ( - c - c ^{2} - 4096 )}{16}} ( 128 + 5 c - 5 c ^{2} - 4096 )}{2 c ^{2} - 4096}

求解步骤

8 y^{''} (t) + c y^{'} (t) + 128 y = 0

求解线性方程:

y = e^{\frac{t ( - c + c ^{2} - 4096 )}{16}} (5 + \frac{128 + 5 c - 5 c ^{2} - 4096}{2 c ^{2} - 4096}) - \frac{e ^{\frac{t ( - c - c ^{2} - 4096 )}{16}} ( 128 + 5 c - 5 c ^{2} - 4096 )}{2 c ^{2} - 4096}

y = e^{\frac{t ( - c + c ^{2} - 4096 )}{16}} (5 + \frac{128 + 5 c - 5 c ^{2} - 4096}{2 c ^{2} - 4096}) - \frac{e ^{\frac{t ( - c - c ^{2} - 4096 )}{16}} ( 128 + 5 c - 5 c ^{2} - 4096 )}{2 c ^{2} - 4096}

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} - \frac{4 y ^{^{'}}}{3} + \frac{97 y}{36} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{31}{2}

4 \frac{d y}{d x} - 3 \frac{d ^{2} y}{d x ^{2}} = 0

y^{^{'''}} + 5 y^{^{''}} - 11 y^{^{'}} - 39 y = 0