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

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

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

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

y = \int - \frac{ln ( e ^{2 x} + 1 )}{2 e ^{- x}} + e^{x} x + c_{1} e^{x} d x

Solution steps

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

Solve using substitution:

y = \int - \frac{ln ( e ^{2 x} + 1 )}{2 e ^{- x}} + e^{x} x + c_{1} e^{x} d x

y = \int - \frac{ln ( e ^{2 x} + 1 )}{2 e ^{- x}} + e^{x} x + c_{1} e^{x} d x

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