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

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

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

y^{^{''}} + 25 y = cos (9 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{2 t}}{e ^{2 t} + 1}

y^{^{''}} - 3 y = e^{2 t} + t e^{2 t} + t^{2} e^{2 t}

y^{^{''}} + 4 y = t^{2} + 8 e^{t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 (9 x - 2) e^{2 x}

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