# DE homework problem

by SomeRandomGuy
Tags: homework
 P: 55 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.
HW Helper
P: 1,021
 Quote by SomeRandomGuy 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.
Could you show your work then? There should be a small difference...
Remember the definition of the absolute value.
P: 55
 Quote by TD Could you show your work then? There should be a small difference... Remember the definition of the absolute value.
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)

 P: 55 DE homework problem 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?
HW Helper
P: 1,021
 Quote by SomeRandomGuy dy/dt = 2*sqrt(abs(y)) = 2*sqrt(y) y>=0, 2*sqrt(-y) y<0
The initial problem was "abs(sqrt(y))" and now you write "sqrt(abs(y))", which one is it?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,569 "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.

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