What is the solution for y^{'''}-y^'=10cos(2x) ?

The solution for y^{'''}-y^'=10cos(2x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x-sin(2x)

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y^{'''}-y^'=10cos(2x)

y^{^{'''}} - y^{^{'}} = 10 cos (2 x)

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

Solution steps

y^{'''} (x) - y^{'} (x) = 10 cos (2 x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{3} - 1

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

x^{^{''}} + 2 x^{^{'}} + 5 x = 8 cos (2 t) + 2 sin (2 t)

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

y^{^{''}} + 16 y = 7 sin (x)

What is the solution for y^{'''}-y^'=10cos(2x) ?
The solution for y^{'''}-y^'=10cos(2x) is y=c_{1}+c_{2}e^{-x}+c_{3}e^x-sin(2x)