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

The solution for y^{''}+16y=2csc^2(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+(-ln(tan(2x))cos(4x)-1)/8

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

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

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- ln ( tan ( 2 x ) ) cos ( 4 x ) - 1}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- ln ( tan ( 2 x ) ) cos ( 4 x ) - 1}{8}

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{- ln ( tan ( 2 x ) ) cos ( 4 x ) - 1}{8}

What is the solution for y^{''}+16y=2csc^2(4x) ?
The solution for y^{''}+16y=2csc^2(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)+(-ln(tan(2x))cos(4x)-1)/8