What is the solution for y^{''}+4y=-2 ?

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

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y^{''}+4y=-2

y^{^{''}} + 4 y = - 2

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

Solution steps

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

Solve linear ODE:

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

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

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