y^'+2y=-8

解答

y^{^{'}} + 2 y = - 8

y = e^{- 2 t + c_{1}} - 4

求解步骤

y^{'} (t) + 2 y = - 8

求解可分离方程:

y = e^{- 2 t + c_{1}} - 4

y = e^{- 2 t + c_{1}} - 4

\frac{d y}{d x} = - \frac{3}{5} x^{7} y^{3}

\frac{d x}{d y} + 2 x y^{2} = 0

\frac{d v}{d x} = - 0.4 v

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 5

(x^{3} + 1) y^{^{'}} = x^{2} y