What is the solution for (-x+2)y^'-y=e^{-2x} ?

The solution for (-x+2)y^'-y=e^{-2x} is y=-(e^{-2x})/(2(-x+2))+(c_{1})/(-x+2)

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(-x+2)y^'-y=e^{-2x}

(- x + 2) y^{^{'}} - y = e^{- 2 x}

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

Solution steps

(- x + 2) y^{'} (x) - y = e^{- 2 x}

Solve linear ODE:

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

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

\int \frac{1}{16 x ^{2} - 4} d x

\frac{\partial}{\partial x} (x^{2} + y^{2} + \frac{32}{x y})

\frac{d}{d x} (sec^{2} (2 x) 2)

x \to 1 lim (e^{(x \cdot 3 - x)})

What is the solution for (-x+2)y^'-y=e^{-2x} ?
The solution for (-x+2)y^'-y=e^{-2x} is y=-(e^{-2x})/(2(-x+2))+(c_{1})/(-x+2)