xy^'+y=7xy^2

解答

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

y = \frac{1}{x ( - 7 ln ( x ) + c _{1} )}

求解步骤

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

求解贝努力方程:

y = \frac{1}{x ( - 7 ln ( x ) + c _{1} )}

y = \frac{1}{x ( - 7 ln ( x ) + c _{1} )}

d er i v a t i v e f (x) = x + 7

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 2)

\frac{d}{d x} (e^{- 2 x} (1 - x))

\frac{d}{d x} (5 ln (x^{5}))

(e^{x})^{^{'}}