-cx^{''}=bx,x(0)=0

解

- c x^{^{''}} = b x, x (0) = 0

x = c_{2} sin (\frac{b}{c} t)

解答ステップ

- c x^{''} (t) = b x

線形 ODE を解く:

x = c_{2} sin (\frac{b}{c} t)

x = c_{2} sin (\frac{b}{c} t)

y^{^{''}} - 2 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 4

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (4) = 2, y^{^{'}} (4) = 6

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 4

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5