Example 1. Verify that the function $y=x+C$ is the general solution of the differential equation $y^{\prime}=1$ and find the particular solution that satisfies the initial condition $\left.y\right|_{x=0}=0$. Provide a geometric interpretation of the result.
Solution. The function $y=x+C$ satisfies the given equation for any value of the arbitrary constant $C$. Indeed, $y'=(x+C)'=1$.

Let us set an arbitrary initial condition $\left.y\right|_{x=x_{0}}=y_{0}$. Substituting $x=x_{0}$ and $y=y_{0}$ into the equation $y=x+C$, we find that $C=y_{0}-x_{0}$. Substituting this value of $C$ into the given function, we get $y=x+y_{0}-x_{0}$. This function satisfies the given initial condition: setting $x=x_{0}$, we get $y=x_{0}+y_{0}-x_{0}=y_{0}$. Thus, the function $y=x+C$ is the general solution of the given equation.

In particular, setting $x_{0}=0$ and $y_{0}=0$, we obtain the particular solution $y=x$.

The general solution of the given equation, i.e., the function $y=x+C$, defines a family of parallel lines in the $x O y$ plane with a slope of $k=1$. Through each point $M_{0}\left(x_{0}, y_{0}\right)$ in the $x O y$ plane, there passes a unique integral curve $y=x+y_{0}-x_{0}$. The particular solution $y=x$ defines one of the integral curves, namely the line passing through the origin (Fig. 4).

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-007.jpg?height=297&width=473&top_left_y=1440&top_left_x=779)

Fig. 4