What is the solution for y^{''}+16y=-16sec^2(4x) ?

The solution for y^{''}+16y=-16sec^2(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1-ln(sec(4x)+tan(4x))sin(4x)

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y^{''}+16y=-16sec^2(4x)

y^{^{''}} + 16 y = - 16 sec^{2} (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 1 - ln (sec (4 x) + tan (4 x)) sin (4 x)

Solution steps

y^{''} (x) + 16 y = - 16 sec^{2} (4 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 1 - ln (sec (4 x) + tan (4 x)) sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + 1 - ln (sec (4 x) + tan (4 x)) sin (4 x)

\frac{\partial}{\partial x} (\frac{c + 1}{c + x})

\frac{d}{d x} (\frac{x - 3}{x + 2})

\frac{d y}{d x} = 3 x y

f (x) = arccos (x^{2})

(0, - 2), (1, 2)

What is the solution for y^{''}+16y=-16sec^2(4x) ?
The solution for y^{''}+16y=-16sec^2(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1-ln(sec(4x)+tan(4x))sin(4x)