What is the solution for y^{''}+y=3csc^2(t) ?

The solution for y^{''}+y=3csc^2(t) is y=c_{1}cos(t)+c_{2}sin(t)-3cos(t)ln(tan(t/2))-3

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y^{''}+y=3csc^2(t)

y^{^{''}} + y = 3 csc^{2} (t)

y = c_{1} cos (t) + c_{2} sin (t) - 3 cos (t) ln (tan (\frac{t}{2})) - 3

Solution steps

y^{''} (t) + y = 3 csc^{2} (t)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - 3 cos (t) ln (tan (\frac{t}{2})) - 3

y = c_{1} cos (t) + c_{2} sin (t) - 3 cos (t) ln (tan (\frac{t}{2})) - 3

\frac{\partial}{\partial x} (\frac{6 y}{x ^{2} + y ^{2} + 5})

\frac{d}{d x} ((4 x - 5)^{3})

\int \frac{1}{e ^{- 3 x}} d x

\frac{d}{d y} (cos (x) y)

\int 25 arctan (x) d x

What is the solution for y^{''}+y=3csc^2(t) ?
The solution for y^{''}+y=3csc^2(t) is y=c_{1}cos(t)+c_{2}sin(t)-3cos(t)ln(tan(t/2))-3