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^{^{''}} + 4 x^{^{'}} + 5 x = 0

\frac{\partial}{\partial n} (nm)

x \to 2 lim (\frac{3 x + 2}{x + 2})

\int_{0}^{2} (2 x + 2 - x^{2} - 2) d x

x \to \infty lim (\frac{ln ( x ^{2} - x + 1 )}{ln ( x ^{10} + x + 1 )})

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)