y^'+y=u,y(0)=(-1)/2

解答

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

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

求解步骤

y^{'} (t) + y = u

求解可分离方程:

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

\frac{d y}{d x} = \frac{- x}{e ^{y}}

\frac{d x}{d t} = 14 - 0.7 x

\frac{d y}{d x} = \frac{3 sin ( x )}{y}

\frac{d y}{d x} + 16 x y = 0

2 x \frac{d y}{d x} + (1 - x) \frac{d y}{d x} - 2 y = 0