What is the solution for y^{''}-y=cosh(t) ?

The solution for y^{''}-y=cosh(t) is y=c_{1}e^t+c_{2}e^{-t}+(e^t(-e^{-2t}+2t)-e^{-t}(2t+e^{2t}))/8

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y^{''}-y=cosh(t)

y^{^{''}} - y = cosh (t)

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

Solution steps

y^{''} (t) - y = cosh (t)

Solve linear ODE:

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

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

d (x) = - 6 x + 4200

y^{^{''}} + 4 y = 4 t^{2} - 4 t + 10, y (0) = 0, y^{^{'}} (0) = 3

y^{^{'''}} - 16 y^{^{'}} = t e^{4 t}

y^{^{''}} + 4 y = 8 t

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- 1}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-y=cosh(t) ?
The solution for y^{''}-y=cosh(t) is y=c_{1}e^t+c_{2}e^{-t}+(e^t(-e^{-2t}+2t)-e^{-t}(2t+e^{2t}))/8