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}

\int cos (φ) d φ

y^{'} = 0.027 y (1 - \frac{y}{1527})

ln (3) + e^{x^{e}}

y^{^{''}} - 5 y^{^{'}} = 10 x + 3

x y y^{^{'}} + x^{2} + y^{2} = 0

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