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

The solution for y^{''}+y^'=cos(2t) is y=c_{1}+c_{2}e^{-t}+1/10 sin(2t)-1/5 cos(2t)

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

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

y = c_{1} + c_{2} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

y = c_{1} + c_{2} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

y^{^{''}} + 16 y = 8 e^{4 t} + 8 sin (2 t)

y^{^{''}} - 6 y^{^{'}} + 15 y = 2 sin (3 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 1, y^{^{'}} (4) = 3

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 36 e^{2 t}

What is the solution for y^{''}+y^'=cos(2t) ?
The solution for y^{''}+y^'=cos(2t) is y=c_{1}+c_{2}e^{-t}+1/10 sin(2t)-1/5 cos(2t)