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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = 8 x^{10} + 6 x^{6}

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

y^{^{'}} + y = - x^{4}

\frac{\partial}{\partial y} (x + \frac{y}{z})

\int 7 x e^{6 x} d x

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