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

y^{^{''}} + y = 6 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = t e^{t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = x - 6

x^{^{''}} + x = cos (3 t)

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))