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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 12 y^{^{'}} + 37 = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} = 2 t^{2} + t - 1

y^{^{''}} - 2 y = 12 x - 10

y^{^{''}} + 2 y^{^{'}} + y = 3 (t - 1) e^{- 1}

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

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