y^'=(3x^2+4x+2)/(2y+1)

解答

y^{^{'}} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

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

求解步骤

y^{'} (x) = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

求解可分离方程:

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

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

(x y) \frac{d y}{d x} = 1 + y^{2}

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

\frac{d y}{d x} x^{2} y - y^{2} = 8

(y ln (x - 2)) y d x = x d y

y^{^{'}} = 6 y^{- 4}, y (0) = 5