What is the solution for y^{'''}-y^'=6sin(t) ?

The solution for y^{'''}-y^'=6sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+3cos(t)

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y^{'''}-y^'=6sin(t)

y^{^{'''}} - y^{^{'}} = 6 sin (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + 3 cos (t)

Solution steps

y^{'''} (t) - y^{'} (t) = 6 sin (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + 3 cos (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + 3 cos (t)

y^{^{''}} + 4 y^{^{'}} + 5 y = (e^{- 2 x}) sec (x)

y^{^{''}} + 6 y^{^{'}} + 8 y = e^{- 2 x} + sin (x)

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

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

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

What is the solution for y^{'''}-y^'=6sin(t) ?
The solution for y^{'''}-y^'=6sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+3cos(t)