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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 2 y = 2 cos (2 x) + sin (2 x)

y^{^{''}} + 4 y = - 5, y (\frac{π}{8}) = - \frac{1}{4}, y^{^{'}} (\frac{π}{8}) = 2

y^{^{''}} + 7 y^{^{'}} + 6 y = 3 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 x^{2} + 12

y^{^{''}} + 2 y^{^{'}} + y = 5 e^{- t}

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