4y^2(xdy+ydx)=ydx-xdy

解答

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

y = \frac{1 + 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}, y = \frac{1 - 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}

求解步骤

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

求解可分离方程:

y = \frac{1 + 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}, y = \frac{1 - 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}

y = \frac{1 + 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}, y = \frac{1 - 1 + 16 e ^{2 c_{1}} x ^{2}}{8 e ^{c_{1}} x}

x y^{2} y^{^{'}} = x + 7

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

y ln (x) \frac{d}{d y} (x) = (\frac{y + 1}{x})^{2}, y = 5, x = 4

y^{^{'}} = \frac{1}{2} x 1 - 16 y^{2}, y (0) = 0

y^{^{'}} = 4 y^{- 2} (x^{6} + x^{4} + x^{2}), y (0) = 0