# Differential equations, qualitative solution

1. Mar 9, 2010

### MaxManus

dy/dt = -ty^2 y(0) = 1
What can you say about the solution over the interval 0<=t <=2?

The text says that the solution is decreasing and never zero because y(t) = 0 for all t is an equilibrium solution.

I can see why the solution can never cross the x-axis, but why can't the solution become zero?

2. Mar 9, 2010

### rock.freak667

What did you get for the solution y(t)?

3. Mar 9, 2010

### MaxManus

I got y = $$\frac{1}{t^2 + 1}$$, but I got the impression that looking at the derivative was enough to make the conclusion

4. Mar 14, 2010

### MaxManus

Anyone?

5. Mar 14, 2010

### rock.freak667

if it crosses the t-axis, then shouldn't y=0 ? and what does that mean?

6. Mar 14, 2010

### MaxManus

Yes, but why can't it become zero and stay there?

7. Mar 14, 2010

### rock.freak667

How exactly would it become zero? That would mean that at some value of 't' the corresponding value of 'y' would be zero.

$$0 = \frac{1}{t^2+1}$$

You would end up with a false equality, which can only mean that y(t) is never zero for all values of t.