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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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