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

The solution for y^{'''}-y^{''}=cos(2t) is y=c_{1}e^t+c_{2}+c_{3}t-1/10 sin(2t)+1/20 cos(2t)

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

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

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{1}{10} sin (2 t) + \frac{1}{20} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{1}{10} sin (2 t) + \frac{1}{20} cos (2 t)

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{1}{10} sin (2 t) + \frac{1}{20} cos (2 t)

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

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{t}, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} - y = 2 t - 4

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = e^{- 3 x} + 8 x^{2}

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{3} e^{3 t}

What is the solution for y^{'''}-y^{''}=cos(2t) ?
The solution for y^{'''}-y^{''}=cos(2t) is y=c_{1}e^t+c_{2}+c_{3}t-1/10 sin(2t)+1/20 cos(2t)