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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (1 + x)

\frac{d}{d x} (\frac{24}{x})

\int_{x}^{x^{2}} e^{t^{4}} d t

\int_{- 1}^{3} (2 x + 3)^{2} d x

\frac{x - y}{x}

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