Example 1. Given the differential equation $y^{\prime}=x$. Construct the direction field and indicate the approximate general solution of this equation.
Solution. We have $f(x, y)=x$. This function is defined for all values of $x$ and $y$ (it does not depend on $y$). This means that $D=\mathbb{R}^{2}$, i.e., the domain $D$ is the entire plane $O x y$. At all points on the plane where $x=0$ (this is the $O y$ axis), we have $y^{\prime}=0$. This means that at points on the $O y$ axis, the field is horizontal (on Fig. 4.1, this is depicted by short horizontal strokes). At points on the plane where $x=1$ (this is the line $x=1$), we have $y^{\prime}=1$, i.e., the tangents to all curves intersecting the line $x=1$ form a $45^{\circ}$ angle with the $O x$ axis. At points on the line $x=-1$, the field (set of tangents) forms a $135^{\circ}$ angle with the positive direction of the $O x$ axis, and so on. The lines $x=K$ are the isoclines of the direction field of the equation, and these directions form angles with the positive direction of the $O x$ axis whose tangents are equal to $K$. Thus, the direction field is constructed. If an integral curve is drawn through any point, say, through the origin $(0,0)$, then at the point $(0,0)$, this curve has a horizontal tangent, and in the neighborhood of the point $x=0$, to the right of it, the slope of the integral curve is positive; it intersects the line $x=1$ at a $45^{\circ}$ angle, the line $x=\sqrt{3}$ at a $60^{\circ}$ angle, the line $x=2$ at a larger angle, and so on. If this integral curve is extended to the left, the slopes of the tangents change such that the line $x=-1$ is intersected at a $135^{\circ}$ angle, the line $x=-\sqrt{3}$ at a $120^{\circ}$ angle, and so forth. Since the field is symmetric with respect to the $O y$ axis, the integral curves will be similar. Similarly, an integral curve passing through another point on the plane, for example, through the point $(0,1)$, can be constructed. The field indicates the "parallelism" of the family of integral curves. A dense network of isoclines and careful construction of integral curves suggest that the integral curves resemble parabolas.

And this is justified, as the general solution of the original equation is $y=\frac{x^{2}}{2}+C$.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-078.jpg?height=480&width=349&top_left_y=1105&top_left_x=2131)

Fig. 4.1