What is the solution for y^{'''}-y^'=2cos(x) ?

The solution for y^{'''}-y^'=2cos(x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x-sin(x)

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y^{'''}-y^'=2cos(x)

y^{^{'''}} - y^{^{'}} = 2 cos (x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - sin (x)

Solution steps

y^{'''} (x) - y^{'} (x) = 2 cos (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - sin (x)

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - sin (x)

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} - 4 y = e^{2 x} - 4 x + 4

y^{^{''}} - y^{^{'}} - 2 y = 15 x + 9

y^{^{''}} - y^{^{'}} - 2 y = 15 x + 8

y^{^{''}} + 6 y^{^{'}} + 25 y = 15 cos (5 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{6} e^{2 t}

What is the solution for y^{'''}-y^'=2cos(x) ?
The solution for y^{'''}-y^'=2cos(x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x-sin(x)