What is the solution for y^{''}+4y=sin(t)-\H(t-2pi)sin(t-2pi) ?

The solution for y^{''}+4y=sin(t)-\H(t-2pi)sin(t-2pi) is

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y^{''}+4y=sin(t)-\H(t-2pi)sin(t-2pi)

y^{^{''}} + 4 y = sin (t) - H (t - 2 π) sin (t - 2 π)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t

Solution steps

y^{''} (t) + 4 y = sin (t) - H (t - 2 π) sin (t - 2 π)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) ( sin ( t ) - H ( t - 2 π ) sin ( t - 2 π ) )}{2} d t

y^{^{''}} + y = sin (a t)

y^{^{''}} - 4 y^{^{'}} + 13 y = 24 e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e^{- t}

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

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

What is the solution for y^{''}+4y=sin(t)-\H(t-2pi)sin(t-2pi) ?
The solution for y^{''}+4y=sin(t)-\H(t-2pi)sin(t-2pi) is