52. Find the particular solution of the equation $2 y d x=(1+x) d y$, if $y=4$ when $x=1$.
Solution. Separating variables:

$$
\frac{2 d x}{1+x}=\frac{d y}{y}
$$

Integrating, we get

$$
\int \frac{2 d x}{1+x}=\int \frac{d y}{y}, 2 \ln (1+x)=\ln y+C, \text { or } \ln (1+x)^{2}=\ln y+\ln C
$$

(here $C$ is replaced by $\ln C$). Exponentiating, we find $(1+x)^{2}=C y$ - the general integral of the given differential equation.

Now let's find the particular solution of the equation given the initial conditions. Substituting $x=1, y=4$ into the general solution, we have $2^{2}=4 C$, from which $C=1$. Therefore, $y=(1+x)^{2}$.