y^'+csc(y)=0

解答

y^{^{'}} + csc (y) = 0

y = arccos (t - c_{1}) + 2 πn, y = - arccos (t - c_{1}) + 2 πn

求解步骤

y^{'} (t) + csc (y) = 0

求解可分离方程:

y = arccos (t - c_{1}) + 2 πn, y = - arccos (t - c_{1}) + 2 πn

y = arccos (t - c_{1}) + 2 πn, y = - arccos (t - c_{1}) + 2 πn

\int \frac{e ^{3 x} + 1}{e ^{x} + 1} d x

x \to 3 lim (2 x^{3} - 4 x - 10)

d er i v a t i v e 9 x^{2} x + 7 x^{2}

s l o p e (- 9, - 2), (3, - 1)

\int (2 x + 1) (x^{2} + x) d x