Example 2. Find the integral curve of the equation $y^{\prime \prime}=x+1$, passing through the point $M_{0}(1,1)$ and tangent to the line $y=\frac{1}{2} x+\frac{1}{2}$ at this point.
Solution. First, we find the general solution of the given equation. We have:

$$
\begin{gathered}
y^{\prime \prime}=\frac{d y^{\prime}}{d x}, \frac{d y^{\prime}}{d x}=x+1, d y^{\prime}=(x+1) d x, y^{\prime}=\int(x+1) d x \\
y^{\prime}=\frac{x^{2}}{2}+x+C_{1} ; \frac{d y}{d x}=\frac{x^{2}}{2}+x+C_{1}, d y=\left(\frac{x^{2}}{2}+x+C_{1}\right) d x \\
y=\frac{x^{3}}{6}+\frac{x^{2}}{2}+C_{1} x+C_{2}
\end{gathered}
$$

We find the particular solution that satisfies the conditions: $y_{0}=1$, $y_{0}^{\prime}=\frac{1}{2}$ at $x_{0}=1$. Substituting these values into the expressions for $y(x)$ and $\boldsymbol{y}^{\prime}(\boldsymbol{x})$, we get:

$$
\left.\left.\begin{array}{c}
y(1)=\frac{1}{6}+\frac{1}{2}+C_{1}+C_{2}=1 ; \\
y^{\prime}(1)=\frac{1}{2}+1+C_{1}=\frac{1}{2},
\end{array}\right\} \begin{array}{c}
C_{1}+C_{2}=\frac{1}{3} \\
C_{1}=-1
\end{array}\right\}
$$

Therefore, $C_{1}=-1, C_{2}=\frac{4}{3}$, so the sought integral curve is defined by the equation

$$
y=\frac{x^{3}}{6}+\frac{x^{2}}{2}-x+\frac{4}{3}
$$