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

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

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x^{''}+4x=cos(4t)

x^{^{''}} + 4 x = cos (4 t)

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

Solution steps

x^{''} (t) + 4 x = cos (4 t)

Solve linear ODE:

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

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

x^{^{''}} - 2 x^{^{'}} + 3 x + 2 = 0

y^{^{''}} - 10 y^{^{'}} + 16 = 0

y^{^{''}} + 2 y^{^{'}} + y = 3 x \cdot e^{- x}

y^{^{'''}} - 12 y^{^{''}} + 44 y^{^{'}} = 48

y^{^{''}} + 1 y^{^{'}} - 20 y = 1200 t^{3}

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