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

The solution for y^{''}+16y=2csc^2(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+(-ln(tan(2t))cos(4t)-1)/8

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (7 x y^{6})

x \to 0 lim (\frac{x - 8}{x - 1})

\int \frac{15}{x ( x ^{2} + 4 ) ^{2}} d x

\int x^{2} (x - 2)^{\frac{3}{2}} d x

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