x^2+2x-1-4yy^'=0

解答

x^{2} + 2 x - 1 - 4 y y^{^{'}} = 0

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

求解步骤

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

求解可分离方程:

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

y = \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}, y = - \frac{x ^{3} + 3 ( x ^{2} - x - c _{1} )}{6}

\frac{d y}{d x} = e^{- 2 a y} - 1

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

y^{^{'}} = 3 t^{2} \frac{y}{y + 1}, y (2) = 1

2 x y + 6 x + (x^{2} + 9) y^{^{'}} = 0

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