129. Form the equation of the curve passing through the point ( $3 ; 4$ ), if the slope of the tangent to this curve at any point ( $x ; y$ ) is equal to $x^{2}-2 x$.
Solution. $1^{0}$. We know that

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-296.jpg?height=369&width=394&top_left_y=1444&top_left_x=747)

Fig. 137
$k=\operatorname{tg} \alpha=\frac{d y}{d x}$. In this case, we have $\frac{d y}{d x}=x^{2}-2 x ; d y=\left(x^{2}-2 x\right) d x$.

As a result of integration, we get $y=\int\left(x^{2}-2 x\right) d x$, i.e.,

$$
y=\frac{1}{3} x^{3}-x^{2}+C
$$

2.. Now from the found family of curves, we will single out the curve that passes through the given point $(3 ; 4)$. For this, we substitute the coordinates of the given point for $x$ and $y$ in equation (1). We have $4=$ $=\frac{1}{3} \cdot 3^{3}-3^{2}+C$, i.e., $C=4$

$3^{0}$. The curve passing through the given point is the one from the family (1) for which $C=4$, i.e., $y=\frac{1}{3} x^{3}-x^{2}+4$.