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

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

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) + ln (cos (2 t)) cos (2 t) + 2 t sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + ln (cos (2 t)) cos (2 t) + 2 t sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + ln (cos (2 t)) cos (2 t) + 2 t sin (2 t)

x y^{^{'}} - y = 2020 x^{2}

(- 7, - 4), (1, - 6)

2 x^{2}, 2 x + 6, - 2, 2

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

\frac{\partial}{\partial θ} (r (θ) sin (θ))

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