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

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

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

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

y = - e^{e^{- x} - c_{1}} + e^{e}

Solution steps

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

Solve separable ODE:

y = - e^{e^{- x} - c_{1}} + e^{e}

y = - e^{e^{- x} - c_{1}} + e^{e}

\int_{0}^{4} x - \frac{1}{2} x d x

f (x) = 4 x - x^{2}, at x = 0

\frac{d}{d x} (1 - cos (\frac{x}{2}))

\frac{d}{d x} (261 e^{0, 5 x} + 27.464)

\frac{d}{d x} (e^{x f (x)})

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