Determine field lines to the vector field V(x, y) = (x, −y)
This is how far I've come. The field line's derivative and the Field function should always be equal for all x & y.
Lets say the field line is $f(x,y)=(x,y)$ and of course we parameterize them and then get $f'(x,y)=(x'(t), y'(t))$. And lets say $u(t)$ is the size of the vector.
$(x'(t), y'(t)) = u(t) (x(t), -y(t)) <=> x'(t)= u(x(t))$ and $y'(t)= -u(t)y(t)$.
After multiplying both equations with x and y, I simplify the entire thing to:
$yx'+xy'=0$.
Have I done right so far? If yes how do I continue solving.
By the way the answer is that the field lines are hyperbolas.
$xy'+x'y =(xy)' =0$
Integrate both sides:
$xy = c $
$y = \dfrac{c}{x} $
Where $c$ is a constant
It is indeed the equation of hyperbola,unless $c =0,$which then will be the x axis