(x+e^y)y^'=xe^{-y}-1

解答

(x + e^{y}) y^{^{'}} = x e^{- y} - 1

y = ln (\frac{- 2 x + 8 x ^{2} + c _{1}}{2}), y = ln (\frac{- 2 x - 8 x ^{2} + c _{1}}{2})

求解步骤

(x + e^{y}) y^{'} (x) = x e^{- y} - 1

求解恰当方程:

y = ln (\frac{- 2 x + 8 x ^{2} + c _{1}}{2}), y = ln (\frac{- 2 x - 8 x ^{2} + c _{1}}{2})

y = ln (\frac{- 2 x + 8 x ^{2} + c _{1}}{2}), y = ln (\frac{- 2 x - 8 x ^{2} + c _{1}}{2})

d er i v a t i v e 2^{x - 1}

x \frac{d y}{d x} + 5 y = x^{3} - x

a re a 7 e^{x}, 7 x e^{x^{2}}, 0, 1

\frac{d}{d x} (2 x^{4} - 5 x^{3} + 4)

h \to 0 lim (- 2 x - h + 3)