What is the solution for 3y^{''}+y=cos(t) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

- y^{^{''}} + y^{^{'}} = 1

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 4 e^{3 t}, y (0) = 2, y^{^{'}} (0) = 7

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