e^yy^'+t/(y+1)= 2/(y+1)

解答

e^{y} y^{^{'}} + \frac{t}{y + 1} = \frac{2}{y + 1}

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

求解步骤

e^{y} y^{'} (t) + \frac{t}{y + 1} = \frac{2}{y + 1}

求解可分离方程:

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

y = W (2 t - \frac{t ^{2}}{2} + c_{1})

x \to \infty lim (\frac{1}{6 ^{x + 1}})

s l o p e (6, 2), (- 3, 5)

d er i v a t i v e \frac{2 x - 3}{4 x + 1}

\int (3 x - 1) x + 6 d x

d er i v a t i v e ln (\frac{2 x + 1}{x ^{2} - 4})