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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} + y = 10 sin (t)

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} + 2 x

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} ln (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 6

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