What is the solution for y^{''}-y=cosh(x) ?

The solution for y^{''}-y=cosh(x) is y=c_{1}e^x+c_{2}e^{-x}+(e^x(-e^{-2x}+2x)-e^{-x}(2x+e^{2x}))/8

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y^{''}-y=cosh(x)

y^{^{''}} - y = cosh (x)

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} ( - e ^{- 2 x} + 2 x ) - e ^{- x} ( 2 x + e ^{2 x} )}{8}

Solution steps

y^{''} (x) - y = cosh (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} ( - e ^{- 2 x} + 2 x ) - e ^{- x} ( 2 x + e ^{2 x} )}{8}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{x} ( - e ^{- 2 x} + 2 x ) - e ^{- x} ( 2 x + e ^{2 x} )}{8}

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

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

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

y^{^{'''}} + y^{^{'}} = sec (t)

y^{^{''}} - 4 y = \frac{e ^{2 x}}{x}

What is the solution for y^{''}-y=cosh(x) ?
The solution for y^{''}-y=cosh(x) is y=c_{1}e^x+c_{2}e^{-x}+(e^x(-e^{-2x}+2x)-e^{-x}(2x+e^{2x}))/8