Example 1. Solve the differential equation $2 x d x-\frac{d y}{\cos ^{2} y}=$ $=0$. Also find the particular solution satisfying the initial condition $y=0$ when $x=0$.
Solution. We have the equation with separated variables: $\frac{d y}{\cos ^{2} y}=2 x d x$. We integrate each part by its own variable: $\int \frac{d y}{\cos ^{2} y}=\int 2 x d x$, i.e., $\operatorname{tg} y=x^{2}+C$. We have obtained the general integral, or the general solution in implicit form. From this, we get the solution in explicit form: $y=\operatorname{arctg}\left(x^{2}+C\right)$ (we do not consider the periodicity of the tangent). Substituting here (or into the general integral) $x=0, y=0$ and determine the value of the constant $C: 0=\operatorname{arctg} C$ (or $0=0+C$). We get $C=0$ and substitute this value into the general solution. The desired particular solution has the form $y=\operatorname{arctg} x^{2}$.

Answer. $y=\operatorname{arctg}\left(x^{2}+C\right) ; y=\operatorname{arctg} x^{2}$.