What is the solution for y^'=-xe^{-x+y} ?

The solution for y^'=-xe^{-x+y} is y=-ln(-e^{-x}x-e^{-x}-c_{1})

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y^'=-xe^{-x+y}

y^{^{'}} = - x e^{- x + y}

y = - ln (- e^{- x} x - e^{- x} - c_{1})

Solution steps

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

Solve separable ODE:

y = - ln (- e^{- x} x - e^{- x} - c_{1})

y = - ln (- e^{- x} x - e^{- x} - c_{1})

\int \frac{( 3 x + 5 )}{x ^{2} + 2 x + 1} d x

\frac{\partial}{\partial x} (x + 4)

\int \frac{- x}{4 + x ^{2}} d x

x \to 5 lim (- x^{2} + 9 x - 7)

\frac{d}{d x} ((3 x^{5} - 1)^{7})

What is the solution for y^'=-xe^{-x+y} ?
The solution for y^'=-xe^{-x+y} is y=-ln(-e^{-x}x-e^{-x}-c_{1})