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

The solution for y^{''}+4^2y=cos(2t) is y=c_{1}cos(4t)+c_{2}sin(4t)+1/12 cos(2t)

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

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

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{12} cos (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{12} cos (2 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{1}{12} cos (2 t)

y^{^{''}} - 10 y^{^{'}} + 21 y = 192 e^{t}

y^{^{''}} - 6 y + 10 y = 10 x^{2}

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

y^{^{''}} + 14 y^{^{'}} + 49 y = x^{- 3} e^{- 7 x}

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

What is the solution for y^{''}+4^2y=cos(2t) ?
The solution for y^{''}+4^2y=cos(2t) is y=c_{1}cos(4t)+c_{2}sin(4t)+1/12 cos(2t)