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# The flow lines (or streamlines) of a vector field are the paths followed by a particle whose velocity field is the given vector field. Thus the vectors in a vector field are tangent to the flow lines. If parametric equations of a flow line are x = x(t), y = y(t) explain why these functions satisfy the differential equations dx/dt=x and dy/dt=-y. Then solve the differential equations to find an equation of the flow line that passes through the point (1, 1).

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(b) Let $x=x(t)$ and $y=y(t)$ the parametric equations of a flow line. By definition of the flow lines, the velocity vector of the flow line at the point $(x, y)$ is is equal to the given vector field.

As the velocity vector is given by $\displaystyle \mathbf{v}=\langle d x(t)/ d t, d y(t) / d t \rangle$, then we have $\mathbf{v}=\mathbf{F}(x,y)$ or

$$$\langle d x(t)/ d t, d y(t) / d t \rangle=\mathbf{F}(x,y)=\langle x, -y\rangle$$$

Therefore, the stream lines satisfy the differential equations

$$$d x / d t=x, \qquad d y / d t=-y$$$

and

$$$d y / d t=-y$$$

To solve (2) we write it as

$d x / x=d t$

Hence, integrating we obtain

$\ln |x|=t+A$

or

$$$x=\pm e^{t+A}=e^{A} e^{t}=C_{1}e^t$$$

where $A$ and $C_{1}$ are constants.

We can solve (3) in a similar way to obtain

$$$y=C_{2} e^{-t}$$$

where $C_{2}$ is a constant.

Taking the product between (4) and (5) we have

$x y=C_{1} e^{t} C_{2} e^{-t}=C_{1}C_{2}=K$

with $K$ a constant.

Therefore, the equations for the flow lines are

$$$y=\frac{K}{x}$$$

If the flow line passes through (1,1) we obtain from (6)

$(1)=\frac{K}{1}$

which gives $K=1$. Then, the flow line through this point has equation

$y=\frac{1}{x},\quad x>0$

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