Example 3. Find the particular solution of the equation $y^{\prime \prime}=y^{\prime} \cdot \cos y$, satisfying the initial conditions $x=1, y=\frac{\pi}{2}, y^{\prime}=1$.
Solution. Visually, we determine that this is an equation of type 3) - it clearly lacks the independent variable $x$. We take $y$ as the new argument and perform the substitution $y^{\prime}=p=p(y(x))$. By the chain rule, we find $y^{\prime \prime}=\left(y^{\prime}\right)_{x}^{\prime}=(p)_{y}^{\prime} \cdot y_{x}^{\prime}=p^{\prime} \cdot p$, and the original equation is reduced to the equation $p \cdot p^{\prime}=p \cos y$, or $p^{\prime}=\cos y$ (here we omit one solution $p=0$, or $y=C$, which is not general: it does not contain two arbitrary constants; if we consider this solution as a particular one, it does not satisfy the initial condition $y^{\prime}=1$). We find $p=\sin y+C$, or $y^{\prime}=\sin y+C$. First, we calculate the value of $C$ based on the initial conditions $y=\frac{\pi}{2}, y^{\prime}=1$. We get $C=0$ and solve the equation $y^{\prime}=\sin y$, i.e., $\frac{d y}{\sin y}=d x$. We find $\ln \left|\operatorname{tg} \frac{y}{2}\right|=x+C_{1}$. From the initial conditions $x=1, y=\frac{\pi}{2}$, we determine $C_{1}=-1$. Then $\operatorname{tg} \frac{y}{2}=e^{x-1}$, i.e., $y=2 \operatorname{arctg} e^{x-1}$.

Answer. $y=2 \operatorname{arctg} e^{x-1}$.

Note. In this example, we needed to find $y$ (a particular solution), so we did not integrate the equation $y^{\prime}=\sin y+C$, or $\frac{d y}{\sin y+C}=d x$ in general form, although this could have been done. This should be taken into account in cases where the intermediate equation is difficult to integrate in general form.