What is the solution for y^{'''}-y^'=3sin(t) ?

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

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y^{'''}-y^'=3sin(t)

y^{^{'''}} - y^{^{'}} = 3 sin (t)

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

Solution steps

y^{'''} (t) - y^{'} (t) = 3 sin (t)

Solve linear ODE:

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

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

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

y^{^{''}} + y^{^{'}} - 6 y = 36 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 20 e^{- t} cos (2 t)

\frac{d ^{2} y}{d x ^{2}} + y = 3 cos (x) - 2 sin (x)

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

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