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

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

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

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

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

Solution steps

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

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

\frac{d}{d x} ((x^{4} + 1)^{\frac{1}{4}})

\frac{d}{d x} (\frac{3 x ^{4} + 7}{x ^{2}})

x \to 0 lim (8)

\int 2 x (x^{2} + 6)^{6} d x

x \to - 0.2 lim (12.1951)

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