y^{''}-(1+2)y^'+21y=0

解答

y^{^{''}} - (1 + 2) y^{^{'}} + 21 y = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2}))

求解步骤

y^{''} (t) - (1 + 2) y^{'} (t) + 21 y = 0

求解线性方程:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2}))

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{5 3 t}{2}) + c_{2} sin (\frac{5 3 t}{2}))

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 5, y^{^{'}} (0) = 22

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = - 5, y^{^{'}} (0) = 29

(d^{4} - 6 d^{3} - 12 d^{2} + 38 d - 21) y = 0

20 y^{^{''}} + 80 y^{^{'}} + 60 y = 0

y^{^{''}} + 5 y^{^{'}} + 25 y = 0