In the 1st lecture of the MIT OCW differential equations course,
at around the 25th minute, the example y’=-x/y is visualized with multiple isoclines, which intersect at (0,0). That is at (0,0) it has multiple (possible?) slopes.
Also if I substitute (0,0) into y'=-x/y or rather y*y'=-x, then I get 0*y’=0, which is true for any slope y’.
But at 37:45, the professor argues that "you can’t have 2 slopes" [at the same point].
Can you resolve this contradiction?
An ordinary differential equation has the form $$y'=f(x,y)\qquad\bigl((x,y)\in\Omega\bigr)\ ,$$ whereby the given function $f$ is defined and continuous in some region, i.e. connected open set $\Omega\subset{\mathbb R}^2$. If $f$is given by some "analytical expression" in the variables $x$ and $y$, as in your example, the tacit understanding is that we should take $\Omega$ as a maximal region where $f$ is defined and continuous. Now $f(x,y):=-{x\over y}$ is undefined when $y=0$. Therefore we can take as $\Omega$ either the upper or the lower half plane of ${\mathbb R}^2$. With both choices $f$ is well defined and even real analytic on $\Omega$. If we take as $\Omega$ the upper half plane the solution curves of the ODE $y'=-{x\over y}$ are the upper semicircles $$\gamma_c:\quad y=\sqrt{c^2-x^2}\qquad (-c<x<c)$$ with $c>0$.