What is the solution for y^{''}+16y=5csc^2(4t) ?

The solution for y^{''}+16y=5csc^2(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+(-5ln(tan(2t))cos(4t)-5)/(16)

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y^{''}+16y=5csc^2(4t)

y^{^{''}} + 16 y = 5 csc^{2} (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 ln ( tan ( 2 t ) ) cos ( 4 t ) - 5}{16}

Solution steps

y^{''} (t) + 16 y = 5 csc^{2} (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 ln ( tan ( 2 t ) ) cos ( 4 t ) - 5}{16}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{- 5 ln ( tan ( 2 t ) ) cos ( 4 t ) - 5}{16}

\int_{0}^{1} π (1 - x^{3})^{2} d x

\frac{\partial}{\partial x} (e^{x} \cdot sin (y))

\int_{0}^{\frac{1}{2}} (4 x + 3) e^{2 x} d x

\frac{\partial}{\partial x} (\frac{5}{x y})

\int \frac{2 x ^{2}}{x ^{2} + 1} d x

What is the solution for y^{''}+16y=5csc^2(4t) ?
The solution for y^{''}+16y=5csc^2(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+(-5ln(tan(2t))cos(4t)-5)/(16)