Example 3. Find the particular solutions of the equation

$$
y^{\prime} \sin x=y \ln y
$$

satisfying the initial conditions:
a) $\left.y\right|_{x=\pi / 2}=e ;$
b) $\left.y\right|_{x=\pi / 2}=1$.
## Solution. We have

$$
\frac{d y}{d x} \sin x=y \ln y
$$

Separating variables

$$
\frac{d y}{y \ln y}=\frac{d x}{\sin x}
$$

Integrating, we find the general integral

$$
\ln |\ln y|=\ln \left|\operatorname{tg} \frac{x}{2}\right|+\ln C
$$

After exponentiation, we get

$$
\ln y=C \cdot \operatorname{tg} \frac{x}{2}, \quad \text { or } \quad y=e^{C \cdot \operatorname{tg}(x / 2)}
$$

which is the general solution of the original equation.

a) Let $x=\frac{\pi}{2}, y=e$, then $e=e^{C \lg (\pi / 4)}$, from which $C=1$. The desired particular solution is $y=e^{\lg (\pi / 2)}$.

b) Assuming in the general solution $x=\frac{\pi}{2}, y=1$, we have $1=e^{C \cdot 18(\pi / 4)}$, from which $C=0$. The desired particular solution is $y \equiv 1$.

Note that in the process of obtaining the general solution, the constant $C$ was under the logarithm, and thus $C=0$ should be considered as a limiting value. This particular solution $y=1$ is contained among the zeros of the product $y \ln y \sin x$, by which we divided both sides of the given equation.