Understanding Solutions of ODEs: y' = -(y^2)

[mt1200]

Hi Physics forums.

I saw this question in a book, I'm not asking for the answer and this is not a homework, I just don't know how to figure out this:

What can you say about a solution of the equation "y' = -(y^2)"just by looking at the differential equation?

I checked at the book's answer and it says " Its either 0 or its decreasing", how did they figured that out?, I see that ODE and I can't think in anything, how do you figure out what an ODE means just by looking at it?.

 

[Integral]

The first deriviative gives you the slope of the solution functions at every point in the xy plane. A good way to get a feeling for what the solutions do is to use the given ODE to find that slope. You can plot a short line with the slope of the solutions though points on the xy plane this will give you a pretty good picture of what the solutions do.

 

[mt1200]

Thanks, that means that as the function's first derivative is negative, its slope its also negative and therefore the function its decrecent!, that makes sense, but how can it be also zero?

 

[tiny-tim]

welcome to pf!

hi mt1200! welcome to pf!

if y is identically zero (y = 0 for all x), then y' = 0 = -0² = -y²

 

[HallsofIvy]

If y is identically 0 then "-y^2" is always 0 so y' is 0 and that just says y is a constant which is consistent with y= 0.

If y is not identically 0, then y^2 is positive, -y^2 is negative, y' is negative so y is decreasing.