Example 4. Find the solution of the equation

$$
(x+y) y^{\prime}=1
$$

satisfying the initial condition: $y=0$ when $x=-1$.
Solution. By appearance, this equation is not linear, as it contains the product of the sought function $y$ and its derivative $y^{\prime}$. However, if we consider $x$ as a function of $y$, then, taking into account that

$$
y^{\prime}=\frac{1}{x^{\prime}}
$$

we obtain the linear equation

$$
x^{\prime}=x+y
$$

Let's apply the substitution

$$
x=u v
$$

then

$$
x^{\prime}=u^{\prime} v+u v^{\prime}
$$

Substituting these expressions into equation (B), we get

$$
u^{\prime} v+u v^{\prime}=u v+y \text { or } u^{\prime} v+u\left(v^{\prime}-v\right)=y \text {. }
$$

Choose the function $v \neq 0$ so that the coefficient of $u$ in the last equation becomes zero

$$
v^{\prime}-v=0, \cdot \frac{d v}{d y}-v=0, \frac{d v}{v}=d y
$$

Integrating, we get

$$
\ln v=y, v=e^{y}
$$

Substituting this value into equation (C), due to equality (D), we find

$$
\begin{gathered}
e^{y} \frac{d u}{d y}=y, d u=y e^{-y} d y \\
u=\int y e^{-y} d y=-y e^{-y}-e^{-y}+C
\end{gathered}
$$

Thus, the general solution of equation (B) is:

$$
x=u v=\left(-y e^{-y}-e^{-y}+C\right) e^{y} \text { or } x=-y-1+C e^{y}
$$

Assuming $x=-1$ and $y=0$, we get $-1=-1+C$, i.e., $C=0$.

Therefore, the sought particular solution is

$$
y=-(x+1)
$$