Find the general solution of differential equation √ (1 + x ^ 2) * sin (2Y) * y '= 2x * sin (y) ^ 2 + e ^ (2 √ (1 + x ^ 2))

Find the general solution of differential equation √ (1 + x ^ 2) * sin (2Y) * y '= 2x * sin (y) ^ 2 + e ^ (2 √ (1 + x ^ 2))

If the right side of the equation is (sin y) ^ 2 instead of sin (y ^ 2), then it can be solved, otherwise it won't be done
Let the root sign (1 + x ^ 2) = Z, DZ / DX = x / Z, - cos (2Y) = g, then dg / DZ = dg / DX * DX / DZ = 2Sin (2Y) * dy / DX * Z / x, then
X / 2 * dg / DZ = 2x * (1 + G) / 2 + e ^ (2Z), that is, DG / DZ = 2G + 2 + 2E ^ (2Z) / root sign (Z ^ 2-1)
(e ^ (- 2Z) g) '= e ^ (- 2Z) * (G' - 2G) = 2 * e ^ (- 2Z) + 2 / radical (Z ^ 2-1)
E ^ (- 2Z) g = - e ^ (- 2Z) + 2ln (Z + radical (Z ^ 2-1)), G (z) = - 1 + 2E ^ (2Z) ln (Z + radical (Z ^ 2-1)). Therefore, the general solution is
G (z) = CE ^ (2Z) - 1 + 2E ^ (2Z) ln (Z + radical (Z ^ 2-1)), that is - cos2y = CE ^ (2 radical (1 + x ^ 2)) - 1 + 2E ^ (2 radical (1 + x ^ 2)) ln (x + radical (1 + x ^ 2))