y^'= 1/2 (k-y)

解答

y^{^{'}} = \frac{1}{2} (k - y)

y = - e^{\frac{- t - c _{1}}{2}} + k

求解步骤

y^{'} (t) = \frac{1}{2} (k - y)

求解可分离方程:

y = - e^{\frac{- t - c _{1}}{2}} + k

y = - e^{\frac{- t - c _{1}}{2}} + k

\frac{d y}{d x} = 6 e^{2 x - y}, y (0) = 0

5 x y d x - ln^{8} (y) d y = 0

y^{^{'}} = \frac{1}{3} y

\frac{d y}{d x} y = u^{2} - 2 u, u = \frac{1}{x ^{2}}

3 = \frac{π}{12} \frac{d}{d t} ((h (t))^{3})