53. Find the particular solution of the differential equation $y^{\prime}=2+y$, if $y=3$ when $x=0$.
Solution. Replacing $y^{\prime}$ with $\frac{d y}{d x}$, and then multiplying all terms by $d x$, we get

$$
d y=2 d x+y d x, \text{ i.e., } d y=(2+y) d x
$$

Dividing both sides of the equation by $2+y$ and integrating:

$$
\frac{d y}{2+y}=d x ; \int \frac{d y}{2+y}=\int d x ; \ln (2+y)=x+\ln C
$$

Express $x$ in terms of the logarithm: $x=\ln e^{x}$. Then we get

$$
\ln (2+y)=\ln e^{x}+\ln C
$$

Exponentiating, we find

$$
2+y=C e^{x}, y=C e^{x}-2
$$

This is the general solution of the given equation.

To find the particular solution, substitute $x=0$, $y=3$ into the general solution and determine $C: 3=C \cdot e^{0}-2 ; e^{0}=1 ; 3=C-2 ; C=5$. Thus, $y=$ $=5 e^{x}-2$.