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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int e^{8 x + 6} d x

y^{^{'}} = \frac{( cos ( 18 x ) )}{( e ^{(18 \cdot y)} )}

x^{2} y^{^{''}} + 3 x y^{^{'}} - 3 y = 0

d y - (y - 9)^{2} d x = 0

x^{\frac{5}{6}}

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