I am given a DE like the following:
$$\frac{dy}{dx} = \frac{2xy}{x^2-1}$$
When one solves it:
$$y = A(x^2-1)$$
Then we obtain an equation for a family of parabolas all intersecting $(-1,0)$ and $(1,0)$.
I did not realise how to solve it so I looked up a solution, and what I had to do is invert the DE on the RHS and multiply it by $-1$ and then find constant by considering the fact that at $(1,1)$ that cruve should be orthogonal to the family of parabolas.
I am quite confused why I would have to swap the denominator and numerator:
$$\frac{dy}{dx} = - \frac{x^2-1}{2xy}$$
I know that if $m$ is slope then orthogonal tangents when multiplied would give $-1$: $m (-1/m) =-1$. And this fact must have something to do with my confusion.
EDIT: To clarify what I was asking there. I am given a DE which gives a family of parabolas when solved. I was then asked to to find a curve that is orthogonal to each of these parabolas at $(1,1) $. And I was confused why in the solution dy/dx was inverted to arrive at such a curve that is orthogonal to all parabolas at $(1,1)$.
I think I have just realised the answer as I was typing. $dy/dx$ is a general equation for the slope at any point in the domain. I have typed up too much now. I shall leave this in case someone else will benefit from this.