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

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

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y^{''}+2y^'+y=(e^x-1)^{-2}

y^{^{''}} + 2 y^{^{'}} + y = (e^{x} - 1)^{- 2}

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} (- x + ln (e^{x} - 1))

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = (e^{x} - 1)^{- 2}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} (- x + ln (e^{x} - 1))

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} (- x + ln (e^{x} - 1))

(D^{2} - 1) y = 2 \cdot (1 + e^{x})^{- 1}

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{5 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = x - sin (x)

y^{^{''}} + 9 y^{^{'}} + 20 = 0

(D^{2} - 4 D + 13) y = 40 sin (3 x)

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