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

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

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

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

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- ln ( tan ( 2 t ) ) cos ( 4 t ) - 1}{16}

Solution steps

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

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- ln ( tan ( 2 t ) ) cos ( 4 t ) - 1}{16}

\int_{1}^{9} \frac{2}{x ^{3}} d x

\int \frac{1}{( 16 t ^{2} + 9 ) ^{2}} d t

\int - 2 e^{3 x} + \frac{5}{x} d x

x \to 0 lim (18 + \frac{20}{x ^{3}})

\int \frac{1}{( 4 x + 3 ) ^{2}} d x

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