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

解

(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})

解答ステップ

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

完全 ODE を解く:

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})