Field lines as solutions to differential equation

Question

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 $F(x, y) = yi + xj$.

Answer 1

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 $y<0$.

Answer 2

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 $\Bbb R^2$.