What is the solution for y^{''}+y=4csc^2(x) ?

The solution for y^{''}+y=4csc^2(x) is y=c_{1}cos(x)+c_{2}sin(x)-4cos(x)ln(tan(x/2))-4

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y^{''}+y=4csc^2(x)

y^{^{''}} + y = 4 csc^{2} (x)

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

Solution steps

y^{''} (x) + y = 4 csc^{2} (x)

Solve linear ODE:

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

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

3 x d y + 2 y d x = 0

\int \frac{3 x ^{3} + 9 x ^{2} + 9 x + 4}{x ^{2} + 3 x + 2} d x

\int (x^{\frac{3}{2}} + 6 x + 5) d x

\frac{d y}{d x} = (x + y)^{2}

- x^{2} - x + 9, - 0.75 x + 6, 0

What is the solution for y^{''}+y=4csc^2(x) ?
The solution for y^{''}+y=4csc^2(x) is y=c_{1}cos(x)+c_{2}sin(x)-4cos(x)ln(tan(x/2))-4