What is the solution for y^{''}+4y=16sec(2t) ?

The solution for y^{''}+4y=16sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4ln(cos(2t))cos(2t)+8tsin(2t)

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y^{''}+4y=16sec(2t)

y^{^{''}} + 4 y = 16 sec (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 ln (cos (2 t)) cos (2 t) + 8 t sin (2 t)

Solution steps

y^{''} (t) + 4 y = 16 sec (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 ln (cos (2 t)) cos (2 t) + 8 t sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 4 ln (cos (2 t)) cos (2 t) + 8 t sin (2 t)

\frac{d}{d t} (4 t^{3})

\int_{0}^{\frac{π}{2}} cos^{5} (x) d x

2 x^{4} + \frac{3 x}{5}, at x = 0

\int (2 + 2 x) e^{(4 x + 2 x^{2})} d x

\int_{0}^{1} \frac{4 x ^{3}}{1 + x ^{4}} d x

What is the solution for y^{''}+4y=16sec(2t) ?
The solution for y^{''}+4y=16sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4ln(cos(2t))cos(2t)+8tsin(2t)