What is the solution for y^'=(x-ysin(2x))/(sin^2(x)) ?

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

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y^'=(x-ysin(2x))/(sin^2(x))

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = 3 sin^{2} (x)

y^{^{''}} + 9 y = t^{2} + cos (3 t)

\frac{s - 5}{s ^{2} + 4 s + 20}

\int \frac{y ^{2}}{( y + 1 ) ^{4}} d y

f (x) = x^{3} + 4 e^{- 0.5 x}

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