What is the solution for y^{''}+4y=-6 ?

The solution for y^{''}+4y=-6 is y=c_{1}cos(2t)+c_{2}sin(2t)-3/2

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y^{''}+4y=-6

y^{^{''}} + 4 y = - 6

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

Solution steps

y^{''} (t) + 4 y = - 6

Solve linear ODE:

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

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

y^{^{''}} + y = sec^{2} (t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 6 e^{6 x}

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

y^{^{''}} - y^{^{'}} - 2 y = 3 x + 4

y^{^{''}} + y = g (t), y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}+4y=-6 ?
The solution for y^{''}+4y=-6 is y=c_{1}cos(2t)+c_{2}sin(2t)-3/2