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

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

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

y^{^{''}} - y^{^{'}} = \frac{1}{e ^{x} + 1}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 e^{5 t}

y^{^{''}} + y = H (t - 1)

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

y^{^{''}} - y^{^{'}} - 6 y = 51 cos (3 x)

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

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