What is the solution for y^{''}+4y=4sec^3(2t) ?

The solution for y^{''}+4y=4sec^3(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tan(2t)sin(2t))/2

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y^{''}+4y=4sec^3(2t)

y^{^{''}} + 4 y = 4 sec^{3} (2 t)

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

Solution steps

y^{''} (t) + 4 y = 4 sec^{3} (2 t)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 24 y = 6 e^{6 x}

y^{^{''}} + 36 y = 4 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 2 y = 3 - 6 t

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 sin (4 t)

y^{^{''}} - 2 y - 3 y^{^{'}} = 3 t^{2} - 5

What is the solution for y^{''}+4y=4sec^3(2t) ?
The solution for y^{''}+4y=4sec^3(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tan(2t)sin(2t))/2