What is the solution for y^{''}+4y=4sec(2x) ?

The solution for y^{''}+4y=4sec(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+ln(cos(2x))cos(2x)+2xsin(2x)

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y^{''}+4y=4sec(2x)

y^{^{''}} + 4 y = 4 sec (2 x)

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

Solution steps

y^{''} (x) + 4 y = 4 sec (2 x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y (π) = π, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + y = x^{2}

y^{^{''}} + y = - 16 cos (3 x)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 7

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} cos (2 x)

What is the solution for y^{''}+4y=4sec(2x) ?
The solution for y^{''}+4y=4sec(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+ln(cos(2x))cos(2x)+2xsin(2x)