y^'+y/4 =4,y(-3)=0

解答

y^{^{'}} + \frac{y}{4} = 4, y (- 3) = 0

y = - 16 e^{\frac{- t - 3}{4}} + 16

求解步骤

y^{'} (t) + \frac{y}{4} = 4

求解可分离方程:

y = - 16 e^{\frac{- t - 3}{4}} + 16

y = - 16 e^{\frac{- t - 3}{4}} + 16

(2 x + 3)^{4} = 3 y^{4} \frac{d y}{d x}

\frac{d y}{d x} = - \frac{4}{6} x^{8} y^{6}

(2 x - 1) y^{^{'}} - 2 y = 0 \frac{1 - 4 x}{x ^{2}}

\frac{d x}{d t} = k (1 - x)

\frac{d y}{d x} + y = y x e^{x + 7}