# Homework Help: DE homework problem

1. Feb 14, 2006

### SomeRandomGuy

Consider dy/dt = 2*(abs(sqrt(y)))

1.Show that y(t)=0 is a solution for all t.
I did this part

2.Find all solutions (hint, give solution like y(t)=... for t>=0, y(t)=... t<0).
He told us in class that t=0 isn't necessarily the point we should be concerned with

3.Why doesn't this contradict the uniqueness theorem?
I have a feeling it's because our DE isn't differentiable at y=0, but my main problem is number 2.

I graphed this DE on the computer, so assuming I typed it in right I know what it looks like. I also tried splitting the DE up into cases for part 2, but it seems that I would have to perform the same integral twice which doesn't really make sense.

2. Feb 14, 2006

### TD

Could you show your work then? There should be a small difference...
Remember the definition of the absolute value.

3. Feb 14, 2006

### SomeRandomGuy

dy/dt = 2*sqrt(abs(y)) = 2*sqrt(y) y>=0, 2*sqrt(-y) y<0

isn't 2*sqrt(-y) when y<0 = 2*sqrt(y)

4. Feb 14, 2006

### SomeRandomGuy

I have y(t) = (t-C)^2 when y>=0. I get the same thing when y<0 as well, by seperation of variables. I use t-C rather than t+C thanks to a hint from my professor from yesterday's lecture. So, is this the solution I am looking for?

5. Feb 15, 2006

### TD

The initial problem was "abs(sqrt(y))" and now you write "sqrt(abs(y))", which one is it?

6. Feb 15, 2006

### HallsofIvy

"isn't 2*sqrt(-y) when y<0 = 2*sqrt(y)"

No, it's not. For example if y= -4, 2*sqrt(-y)= 2*sqrt(4)= 4 but
2*sqrt(y)= 2*sqrt(-4)= 4i.