What is the solution for y^{''}+y= 1/(sin^2(x)) ?

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

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y^{''}+y= 1/(sin^2(x))

y^{^{''}} + y = \frac{1}{sin ^{2} ( x )}

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

Solution steps

y^{''} (x) + y = \frac{1}{sin ^{2} ( x )}

Solve linear ODE:

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

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

5 y^{^{''}} + 10 y^{^{'}} = 3 x^{2}

x^{^{''}} + 4 x^{^{'}} + 3 x = sin (e^{t})

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

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

y^{^{''}} + 4 y + 4 y = (3 + x) e^{- 2 x}

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