dy/dx - y/x = 0
I got 1/x, but I am not sure if this is right or not.
well, did you try it?
d/dx (y/x) = 1/x y' - y/x^2
so, now you have an exact differential, which is what you wanted.
I got 1/x, but I am not sure if this is right or not.
d/dx (y/x) = 1/x y' - y/x^2
so, now you have an exact differential, which is what you wanted.