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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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