What is the solution for y^{''}-0.16y=32cosh(0.4x) ?

The solution for y^{''}-0.16y=32cosh(0.4x) is

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y^{''}-0.16y=32cosh(0.4x)

y^{^{''}} - 0.16 y = 32 cosh (0.4 x)

y = c_{1} e^{\frac{2 x}{5}} + c_{2} e^{- \frac{2 x}{5}} + 50 e^{\frac{2 x}{5}} (- \frac{1}{2} e^{- \frac{4 x}{5}} + \frac{2 x}{5}) - 50 e^{- \frac{2 x}{5}} (\frac{2 x}{5} + \frac{1}{2} e^{\frac{4 x}{5}})

Solution steps

y^{''} (x) - 0.16 y = 32 cosh (0.4 x)

Solve linear ODE:

y = c_{1} e^{\frac{2 x}{5}} + c_{2} e^{- \frac{2 x}{5}} + 50 e^{\frac{2 x}{5}} (- \frac{1}{2} e^{- \frac{4 x}{5}} + \frac{2 x}{5}) - 50 e^{- \frac{2 x}{5}} (\frac{2 x}{5} + \frac{1}{2} e^{\frac{4 x}{5}})

y = c_{1} e^{\frac{2 x}{5}} + c_{2} e^{- \frac{2 x}{5}} + 50 e^{\frac{2 x}{5}} (- \frac{1}{2} e^{- \frac{4 x}{5}} + \frac{2 x}{5}) - 50 e^{- \frac{2 x}{5}} (\frac{2 x}{5} + \frac{1}{2} e^{\frac{4 x}{5}})

y^{^{''}} + 9 y = e^{4 x} + 3 x^{2} - 5

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 3 e^{x} sec (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = 10 e^{t} - 250 cos (3 t)

What is the solution for y^{''}-0.16y=32cosh(0.4x) ?
The solution for y^{''}-0.16y=32cosh(0.4x) is