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

The solution for y^{''}+y=cos((-t+pi)/2) is y=c_{1}cos(t)+c_{2}sin(t)+4/3 cos((-t+pi)/2)

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

y^{^{''}} + y = cos (\frac{- t + π}{2})

y = c_{1} cos (t) + c_{2} sin (t) + \frac{4}{3} cos (\frac{- t + π}{2})

Solution steps

y^{''} (t) + y = cos (\frac{- t + π}{2})

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + \frac{4}{3} cos (\frac{- t + π}{2})

y = c_{1} cos (t) + c_{2} sin (t) + \frac{4}{3} cos (\frac{- t + π}{2})

y^{^{''}} - y^{^{'}} - 6 y = 2 sin (3 x)

(D^{2} + 9) y = 4 cos (3 x)

y^{^{''}} + 2 y^{^{'}} = 2 x + 9 - e - 2 x

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

y^{^{''}} - 12 y^{^{'}} + 36 y = (x + 1) e^{x}

What is the solution for y^{''}+y=cos((-t+pi)/2) ?
The solution for y^{''}+y=cos((-t+pi)/2) is y=c_{1}cos(t)+c_{2}sin(t)+4/3 cos((-t+pi)/2)