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

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

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

y^{^{''}} + 4 y = \frac{4}{cos ( 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 = \frac{4}{cos ( 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)

\int - t e^{- t} d t

\int (8 + x^{3})^{5} (3 x^{2}) d x

f (x) = \frac{1}{5 x ^{2} + 4 x}

\int_{1}^{2} (6 x^{2} - 3 x) d x

\frac{\partial}{\partial x} (x y - x)

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