Fluid dynamics computing streamlines I have a question where I have to compute the streamlines given the fluid flow $\vec{u} = (y,x)$. I must show the streamlines are $x^2 -y^2 = \;$ const. I start of by writing some differential statements: 
$\frac{dx}{dt} = y \; \; \; \;$ as well as $\; \; \; \; \frac{dy}{dt} = x$
Then I isolate for $dt$:
$\rightarrow dt = \frac{dx}{y}  \; \; \; \;$ as well as $ \; \; \; \; \rightarrow dt = \frac{dy}{x} $ 
Then I simply eliminate $dt$ altogether by substituting one of the equations into the other and obtaining a simple separable first order ODE:
$\rightarrow x\:dx = y \:dy$
eventually leading to:
$\rightarrow x^2 - y^2 =\; $Const
as was required in the problem, and I am satisfied with my final answer however I am unsure if my move of eliminating $dt$ through substitution is valid. If not can someone please explain why not. Thanks for all your time!
 A: Well, manipulating differentials such as $dt$ as if they were subject to the field axioms of $\Bbb R$ or $\Bbb C$ is always a little bit dicey, since they are not numbers per se.  However, a rigorous derivation of the equation
$x^2 - y^2 = \text{Const} \tag 1$
may be obtained, assuming that
$\dfrac{dx}{dt} \ne 0, \tag 2$
by using the relationship
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}; \tag 3$
with the given equations
$\dfrac{dx}{dt} = y, \tag 4$
$\dfrac{dy}{dt} = x, \tag 5$
we may (assuming for the moment that $dx/dt \ne 0$, that is, that $y \ne 0$) write (3) as
$\dfrac{dy}{dx} = \dfrac{x}{y}; \tag 6$
then
$y \dfrac{dy}{dx} = x, \tag 7$
so we may integrate to find
$\dfrac{1}{2}y^2 = \displaystyle \int y \; dy = \int y\dfrac{dy}{dx} \; dx = \int x \; dx = \dfrac{1}{2}x^2 + c, \tag 8$
or
$x^2 - y^2 = C = -2c. \tag 9$
If we wish to solve in a region in which $dx/dt = y = 0$, we can always reciprocate (3) (as long as $dy/dt = x \ne 0)$ and use
$\dfrac{dx}{dy} = \dfrac{dx/dt}{dy/dt}, \tag{10}$
leading to
$\dfrac{dx}{dy} = \dfrac{y}{x}, \tag{11}$
which gives the same result.
A: What you are doing here, is an abuse of notation but it happens to work out. It’s similar to how multiplying by $dx$ and other techniques are often used to solve separable odes.
If you want to be more mathematically rigorous you can say,
$$\frac{dx}{dt} \frac{dy}{dx}=\frac{dy}{dt}$$
By the chain rule. So,
$$y \frac{dy}{dx}=x$$
A seperable ode.  That can be solved, more rigorously, by integrating both sides with respect to $x$ and using the chain rule in reverse.
$$ \int y \frac{dy}{dx} dx= \int x dx$$
$$\frac{y^2}{2}=\frac{x^2}{2}+C_1$$
