y^'=-(3x+3xy^2)/(yx^2+2y)

解答

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

y = \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}, y = - \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}

求解步骤

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

求解可分离方程:

y = \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}, y = - \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}

y = \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}, y = - \frac{c _{1} - x ^{6} - 6 x ^{4} - 12 x ^{2} - 8}{x ^{6} + 6 x ^{4} + 12 x ^{2} + 8}

y^{2} d x + x^{2} d y = 0

\frac{d y}{d x} y = ln (ln (2 x))

9 y \frac{d y}{d x} = 2 x^{2}

x y^{2} d x = e^{x} d y

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