# 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. 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. Find the field lines of F(x,y)=yi+xj.

Instead of using arrows to represent a planar vector field, one sometimes uses families of curves called field lines. A curve $y=y\left(x\right)$ is a field line of the vector field F(x,y) if at each point $\left({x}_{0},{y}_{0}\right)$ on the curve, $F\left({x}_{0},{y}_{0}\right)$ is tangent to the curve.
Show that the field lines y=y(x) of a vector field $F\left(x,y\right)=P\left(x,y\right)i+Q\left(x,y\right)j$ are solutions to the differential equation $dy/dx=Q/P$
Find the field lines of $F\left(x,y\right)=yi+xj$
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Branson Perkins
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 $\alpha \left(x\right)$ such that
$\left(P\left(x,y\left(x\right)\right),Q\left(x,y\left(x\right)\right)\right)=\alpha \left(x\right)\phantom{\rule{thinmathspace}{0ex}}\left(1,{y}^{\prime }\left(x\right)\right).$
Thus

Thus
${y}^{\prime }\left(x\right)=\frac{Q\left(x,y\left(x\right)\right)}{P\left(x,y\left(x\right)\right)}.$
When
$F\left(x,y\right)=\left(y,x\right),$
the differential equation becomes
${y}^{\prime }=\frac{x}{y},$
with solution $y=\sqrt{{x}^{2}+y\left({x}_{0}{\right)}^{2}-{x}_{0}^{2}}$ when y>0 and $y=-\sqrt{{x}^{2}+y\left({x}_{0}{\right)}^{2}-{x}_{0}^{2}}$ when y<0.
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charlygyloavao9
We consider the curve
$\begin{array}{}\text{(1)}& \alpha \left(x\right)=\left(x,y\left(x\right)\right),\end{array}$
with tangent vector
$\begin{array}{}\text{(2)}& {\alpha }^{\prime }\left(x\right)=\left(1,{y}^{\prime }\left(x\right)\right)=\mathbf{i}+{y}^{\prime }\left(x\right)\mathbf{j};\end{array}$
if
$\begin{array}{}\text{(3)}& F\left(x,y\right)=P\left(x,y\right)\mathbf{i}+Q\left(x,y\right)\mathbf{j}\end{array}$
is tangent to $\alpha \left(x\right)$ at (x,y), then ${\alpha }^{\prime }\left(x\right)$ is collinear with F(x,y); that is, there is some
$\begin{array}{}\text{(4)}& 0\ne \beta \in \mathbb{R}\end{array}$
with
$\begin{array}{}\text{(5)}& {\alpha }^{\prime }\left(x\right)=\beta F\left(x,y\right);\end{array}$
that is, by virtue of (2) and (3),
$\begin{array}{}\text{(6)}& \mathbf{i}+{y}^{\prime }\left(x\right)\mathbf{j}=\beta P\left(x,y\right)\mathbf{i}+\beta Q\left(x,y\right)\mathbf{j};\end{array}$
comparing coefficients yields
$\begin{array}{}\text{(7)}& \beta P\left(x,y\right)=1,\end{array}$
and
$\begin{array}{}\text{(8)}& \beta Q\left(x,y\right)={y}^{\prime }\left(x\right);\end{array}$
we observe that (7) implies $P\left(x,y\right)\ne 0$, hence we have
$\begin{array}{}\text{(9)}& \beta =\frac{1}{P\left(x,y\right)},\end{array}$
and combining this with (8) we find
$\begin{array}{}\text{(10)}& {y}^{\prime }\left(x\right)=\beta Q\left(x,y\right)=\frac{1}{P\left(x,y\right)}Q\left(x,y\right)=\frac{Q\left(x,y\right)}{P\left(x,y\right)}.\end{array}$
Now with
$\begin{array}{}\text{(11)}& F\left(x,y\right)=y\mathbf{i}+x\mathbf{j},\end{array}$
we obtain
$\begin{array}{}\text{(12)}& {y}^{\prime }\left(x\right)=\frac{x}{y},\end{array}$
or
$\begin{array}{}\text{(13)}& y{y}^{\prime }\left(x\right)=x;\end{array}$
we observe that
$\begin{array}{}\text{(14)}& \frac{1}{2}\left({y}^{2}\left(x\right){\right)}^{\prime }=y{y}^{\prime }\left(x\right);\end{array}$
(13) may thus be written as
$\begin{array}{}\text{(15)}& \frac{1}{2}\left({y}^{2}\left(x\right){\right)}^{\prime }=\frac{1}{2}\left({x}^{2}{\right)}^{\prime },\end{array}$
or
$\begin{array}{}\text{(16)}& \frac{1}{2}\left({y}^{2}\left(x\right)-{x}^{2}{\right)}^{\prime }=0,\end{array}$
whence
$\begin{array}{}\text{(17)}& \left({y}^{2}\left(x\right)-{x}^{2}{\right)}^{\prime }=0,\end{array}$
which implies that
$\begin{array}{}\text{(18)}& {y}^{2}\left(x\right)-{x}^{2}=C,\phantom{\rule{thickmathspace}{0ex}}\text{a constant};\end{array}$ a constant;(18)
the field lines of (11) are thus the curves
$\begin{array}{}\text{(19)}& {y}^{2}-{x}^{2}=C,\end{array}$
which is a family of hyperbolas in ${\mathbb{R}}^{2}$