Example 2. Solve the equation $y^{\prime}=\frac{1}{x \cos y+\sin 2 y}$.
Solution. The given equation is not linear (or a Bernoulli equation), so a substitution of the form $y=u v$ is not applicable. It is necessary to recall the properties of inverse functions and the relationship between their derivatives: $x^{\prime} \cdot y^{\prime}=1, \quad y^{\prime}=\frac{1}{x^{\prime}}$, to replace the given equation with the equation $x^{\prime}=x \cos y+\sin 2 y$, which is linear with respect to the function $x=x(y)$ of the argument $y$.

We perform the substitution $x=u \cdot v, \quad x^{\prime}=u^{\prime} v+u v^{\prime}$. Substituting into the new equation: $u^{\prime} v+u v^{\prime}=u \cdot v \cos y+\sin 2 y$, after which we can use the scheme applied in the previous example.

1) We equate the middle terms of the equation: $u v^{\prime}=u v \cdot \cos y, \quad v^{\prime}=$ $=v \cos y, \quad \ln |v|=\sin y, \quad v=e^{\sin y}$ (the constant of integration is taken to be zero).
2) We are left with the equality of the extreme terms: $u^{\prime} v=\sin 2 y$, $u^{\prime} \cdot e^{\sin y}=\sin 2 y, u^{\prime}=2 \sin y \cos y e^{-\sin y}, u=2 \int \sin y \cos y e^{-\sin y} d y=$ $=-2(1+\sin y) e^{-\sin y}+C$ (integrated by parts). Multiplying by $v$, we obtain the expression for $x=x(y)$.

Answer. $x=C e^{\sin y}-2(1+\sin y)$.