y^'=(sin(2x))/(cos(3y)),y(0)=1

解答

y^{^{'}} = \frac{sin ( 2 x )}{cos ( 3 y )}, y (0) = 1

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

求解步骤

y^{'} (x) = \frac{sin ( 2 x )}{cos ( 3 y )}

求解可分离方程:

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

y = \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) + sin ( 3 ) + \frac{3}{2} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( \frac{3}{2} cos ( 2 x ) - sin ( 3 ) - \frac{3}{2} )}{3} + \frac{2 πn}{3}

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

y^{^{'}} + y = y (t e^{t^{2}} + 1), y (0) = 1

y^{^{'}} = \frac{y ^{2}}{x ^{3}}, y (2) = 4

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

(3 y^{2} - 4 x^{- 1} y) d x + (3 x y - 4) d y = 0