# Field lines as solutions to differential equation

Instead of using arrows to represent a planar vector field, one sometimes uses families of curves called field lines. A curve $$y = y(x)$$ is a field line of the vector field $$F(x, y)$$ if at each point $$(x_0, y_0)$$ on the curve, $$F(x_0, y_0)$$ is tangent to the curve.

1. Show that the field lines $$y = y(x)$$ of a vector field $$F(x,y) = P(x,y)i + Q(x,y)j$$ are solutions to the differential equation $$dy/dx = Q/P$$.
2. Find the field lines of $$F(x, y) = yi + xj$$.

If $$y(x)$$ is a field line, this means that at every $$x$$ you have that $$F(x,y)$$ is colinear with the derivative of $$(x,y(x))$$, which is $$(1,y'(x))$$. So for each $$x$$ there is a number $$\alpha(x)$$ such that $$(P(x,y(x)),Q(x,y(x)))=\alpha(x)\,(1,y'(x)).$$ Thus $$P(x,y(x))=\alpha(x),\ \ \ Q(x,y(x))=\alpha(x)\,y'(x)=P(x,y(x))\,y'(x).$$ Thus $$y'(x)=\frac{Q(x,y(x))}{P(x,y(x))}.$$ When $$F(x,y)=(y,x),$$ the differential equation becomes $$y'=\frac{x}{y},$$ with solution $$y=\sqrt{x^2+y(x_0)^2-x_0^2}$$ when $$y>0$$ and $$y=-\sqrt{x^2+y(x_0)^2-x_0^2}$$ when $$y<0$$.

We consider the curve

$$\alpha(x) = (x, y(x)), \tag 1$$

with tangent vector

$$\alpha'(x) = (1, y'(x)) = \mathbf i + y'(x) \mathbf j; \tag 2$$

if

$$F(x, y) = P(x, y) \mathbf i + Q(x, y) \mathbf j \tag 3$$

is tangent to $$\alpha(x)$$ at $$(x, y)$$, then $$\alpha'(x)$$ is collinear with $$F(x, y)$$; that is, there is some

$$0 \ne \beta \in \Bbb R \tag 4$$

with

$$\alpha'(x) = \beta F(x, y); \tag 5$$

that is, by virtue of (2) and (3),

$$\mathbf i + y'(x) \mathbf j = \beta P(x, y) \mathbf i + \beta Q(x, y) \mathbf j; \tag 6$$

comparing coefficients yields

$$\beta P(x, y) = 1, \tag 7$$

and

$$\beta Q(x, y) = y'(x); \tag 8$$

we observe that (7) implies $$P(x, y) \ne 0$$, hence we have

$$\beta = \dfrac{1}{P(x, y)}, \tag 9$$

and combining this with (8) we find

$$y'(x) = \beta Q(x, y) = \dfrac{1}{P(x, y)} Q(x, y) = \dfrac{Q(x, y)}{P(x, y)}. \tag{10}$$

Now with

$$F(x, y) = y \mathbf i + x \mathbf j, \tag{11}$$

we obtain

$$y'(x) = \dfrac{x}{y}, \tag{12}$$

or

$$yy'(x) = x; \tag{13}$$

we observe that

$$\dfrac{1}{2}(y^2(x))' = yy'(x); \tag{14}$$

(13) may thus be written as

$$\dfrac{1}{2}(y^2(x))' = \dfrac{1}{2}(x^2)', \tag{15}$$

or

$$\dfrac{1}{2}(y^2(x) - x^2)' = 0, \tag{16}$$

whence

$$(y^2(x) - x^2)' = 0, \tag{17}$$

which implies that

$$y^2(x) - x^2 = C, \; \text{a constant}; \tag{18}$$

the field lines of (11) are thus the curves

$$y^2 - x^2 = C, \tag{19}$$

which is a family of hyperbolas in $$\Bbb R^2$$.