-x/y =y^'

解答

- \frac{x}{y} = y^{^{'}}

y = - x^{2} - c_{1}, y = - - x^{2} - c_{1}

求解步骤

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

求解可分离方程:

y = - x^{2} - c_{1}, y = - - x^{2} - c_{1}

y = - x^{2} - c_{1}, y = - - x^{2} - c_{1}

y^{^{'}} = \frac{- x y ^{2}}{4}, y (0) = 2

\frac{d y}{d t} - 4 y = 0

(t^{2} - 14 t + 45) (\frac{d y}{d t}) = y

(1 + y^{2}) \cdot cos (x) \cdot d x - \frac{1}{x} \cdot d y = 0

a y^{^{'}} = - b y^{2} - c