60. Find the equation of the line passing through the point $(1 ; 3)$ and having a tangent with a slope equal to $2 x-3$ (see problem 19).
Solution. Using the condition, we form the differential equation:

$$
y^{\prime}=2 x-3 ; \frac{d y}{d x}=2 x-3 ; d y=(2 x-3) d x
$$

Let's find the general solution of this equation:

$$
\int d y=\int(2 x-3) d x ; y=x^{2}-3 x+C
$$

Substituting the initial data $x=1, y=3$ into the general solution, we get $C=5$. Therefore, the particular solution has the form $y=x^{2}-3 x+5$.

The family of integral curves corresponding to the general solution of the differential equation are parabolas $y=x^{2}-3 x+C$ with vertices located on the line $x=2 / 3$. The found particular solution corresponds to a parabola passing through the point ( $1 ; 3$ ) and intersecting the $O y$ axis at the point $(0 ; 5)$.