What is the solution for y^{'''}-y^'=2*cos(t) ?

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

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y^{'''}-y^'=2*cos(t)

y^{^{'''}} - y^{^{'}} = 2 \cdot cos (t)

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

Solution steps

y^{'''} (t) - y^{'} (t) = 2 \cdot cos (t)

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (2 t)

y^{^{''}} - 3 y^{^{'}} - 6 y = 2 t^{2} + 2 t - 5 e^{3 t}

d (x) = 2 x - 4

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 sin (2 x) - 3 e^{2} x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = e^{x}

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