What is the solution for y^{''}+4y=cos(t) ?

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

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y^{''}+4y=cos(t)

y^{^{''}} + 4 y = cos (t)

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

Solution steps

y^{''} (t) + 4 y = cos (t)

Solve linear ODE:

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

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

y^{^{'''}} + 5 y^{^{''}} + 6 y^{^{'}} = 4 + 9 sin (x)

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

y^{^{''''}} + 9 y^{^{''}} = cos (3 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{3 t}

y^{^{''}} + y = 1 - H (t - 3 π)

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