x^{''}=-2x^'

解

x^{^{''}} = - 2 x^{^{'}}

x = c_{1} + c_{2} e^{- 2 t}

解答ステップ

x^{''} (t) = - 2 x^{'} (t)

線形 ODE を解く:

x = c_{1} + c_{2} e^{- 2 t}

x = c_{1} + c_{2} e^{- 2 t}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (1) x = e^{3 x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 4

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

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