What is the solution for y^{''}+2y=cos(8t) ?

The solution for y^{''}+2y=cos(8t) is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-1/62 cos(8t)

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y^{''}+2y=cos(8t)

y^{^{''}} + 2 y = cos (8 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{62} cos (8 t)

Solution steps

y^{''} (t) + 2 y = cos (8 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{62} cos (8 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{62} cos (8 t)

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

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

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

y^{^{''}} + 9 y = sin (x) + cos (x)

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

What is the solution for y^{''}+2y=cos(8t) ?
The solution for y^{''}+2y=cos(8t) is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-1/62 cos(8t)