y^'=(2cos(2x))/(3+2y)

解答

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

y = \frac{- 3 + 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}

求解步骤

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

求解可分离方程:

y = \frac{- 3 + 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}

y = \frac{- 3 + 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}, y = \frac{- 3 - 9 - 4 ( - c _{1} - sin ( 2 x ) )}{2}

(y^{2} + y) - x \frac{d y}{d x} = 0

y^{^{'}} - 3 y^{^{'}} + 2 y = 0

\frac{d y}{d x} (3 x^{2} y) = 5 x^{2}

y^{^{'}} = 3 x^{2} (y^{2} + 16), y (0) = 0

x^{2} y^{2} + (1 + y^{2}) y^{^{'}} = 0