What is the solution for 2y^{''}+y=5+cos(4t) ?

The solution for 2y^{''}+y=5+cos(4t) is y=c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2)))+5-1/31 cos(4t)

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2y^{''}+y=5+cos(4t)

2 y^{^{''}} + y = 5 + cos (4 t)

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + 5 - \frac{1}{31} cos (4 t)

Solution steps

2 y^{''} (t) + y = 5 + cos (4 t)

Solve linear ODE:

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + 5 - \frac{1}{31} cos (4 t)

y = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + 5 - \frac{1}{31} cos (4 t)

y^{^{''}} + 4 y^{^{'}} + 3 y = x - 1

y^{^{'''}} - 6 y^{^{''}} = cos (4 x)

y^{^{''}} - 2 y^{^{'}} - 8 y = - 8 t - 18, y (0) = 1, y^{^{'}} (0) = 28

y^{^{''}} - 2 y^{^{'}} + y = e^{x} sec^{2} (x) tan (x)

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

What is the solution for 2y^{''}+y=5+cos(4t) ?
The solution for 2y^{''}+y=5+cos(4t) is y=c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2)))+5-1/31 cos(4t)