y^'=(-0.8)/(cos(y)),y(0)= pi/2

解答

y^{^{'}} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

求解步骤

y^{'} (t) = \frac{- 0.8}{cos ( y )}

求解可分离方程:

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

y = arcsin (- 0.8 t + 1) + 2 πn, y = π + arcsin (0.8 t - 1) + 2 πn

f^{^{'}} (x) = f (x) (x + 1)

y^{^{'}} = 0.2877 y, y (0) = 150

y^{^{'}} = 5 x y^{2}

y^{^{'}} = x^{2} y

y^{^{'}} - \frac{x}{5 y ^{4}} = 0, y (0) = 2