Example 4.5. Find a particular solution of the equation $\left(1+e^{x}\right) y y^{\prime}=\mathrm{e}^{x}$, satisfying the initial condition $\left.y\right|_{x=0}=1$.
Solution. We have $\left(1+e^{x}\right) y \frac{d y}{d x}=e^{x}$. Separating variables, we get: $y d y=\frac{e^{x} d x}{1+e^{x}}$. Integrating, we find the general integral:

$$
\frac{y^{2}}{2}=\ln \left(1+e^{x}\right)+C
$$

Now let's find $C$. Setting $x=0, y=1$, we get

$$
\frac{1}{2}=\ln \left(1+e^{0}\right)+C \Rightarrow C=\frac{1}{2}-\ln 2
$$

Substituting the value of $C$ into the expression for the general integral, we find the particular integral:

$$
\frac{y^{2}}{2}=\ln \left(1+e^{x}\right)+\frac{1}{2}-\ln 2
$$

or

$$
y^{2}=1+2 \ln \frac{1+e^{x}}{2}=1+\ln \frac{1+e^{x^{2}}}{2},
$$

from which

$$
y=+\sqrt{1+\ln \frac{1+e^{x^{2}}}{2}}
$$

From the initial condition, it follows that $y>0\left(\left.y\right|_{x=0}=1\right)$, so we take the positive sign before the root. Thus, the required particular solution is:

$$
y=\sqrt{1+\ln \frac{1+e^{x^{2}}}{2}} \quad \text { or } x=\ln \left(2 e^{\frac{y^{2}-1}{2}}-1\right)
$$