Hi again PG,
Yes, that's good. We can condense plus and minus c to just c because after all it's just a constant and can take on positive or negative values. That leaves us with:
y=c*x (or y=c*x^3 in your case)
and
y=c*abs(x) (or y=c*abs(x^3) in your case)
We're down to two possible solutions now, so the next step would be to insert each one into the original diff equation and see if one or both or none works.
I'd like to show you this and then you can do it with your case and see what you can find out.
So we'll look at the dy/dx=y/x first, with possible solutions y=c*x and y=c*abs(x).
Im going to use the symbols "!=" for "not equal to".
For y=c*x we have dy/dx=c and y/x=c, so that must be a solution because dy/dx=y/x.
For y=c*|x| we have dy/dx=c*signum(x) and y/x=c*signum(x), but there's a catch here because dy/dx!=y/x for all x. In fact, dy/dx!=y/x for x=0, so this may invalidate the solution unless we include that exception with the solution.
dy/dx!=y/x for x=0 because lim x-->0+ (dy/dx) != lim x-->0- (dy/dx).
So we can write:
y=c*x
as a solution, but we cant write:
y=c*|x|
as a solution unless we also include the exception like this:
y=c*|x|, x!=0
Most textbooks present the result of dy/dx=y/x as simply y=c*x, but textbooks sometimes dont include an entire discussion that would be necessary. A solution in some forums is a solution that fits for *all* x, but in fact a solution only has to be valid over the range in which it is specified in real life. If you are working a course you'd have to find out what the instructor expects to get a good mark on this
Now it's your turn to do your original problem again and check for any inconsistencies.