What is the solution for y^{''}=-4y-cos(3) ?

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

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y^{''}=-4y-cos(3)

y^{^{''}} = - 4 y - cos (3)

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

Solution steps

y^{''} (t) = - 4 y - cos (3)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = a

(D^{2} + 3 D + 2) y = \frac{1}{4 + e ^{x}}

y^{^{''}} + 2 y^{^{'}} - 2 y = 2 x + 6 + e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 6, y (0) = 6, y^{^{'}} (0) = 17

y^{^{''}} - 8 y^{^{'}} + 16 y = (7 t^{4} - 2 t^{3}) e^{4 t}

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