
#1
Feb1406, 02:51 PM

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. 



#2
Feb1406, 03:03 PM

HW Helper
P: 1,024

Remember the definition of the absolute value. 



#3
Feb1406, 04:47 PM

P: 55

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



#4
Feb1406, 07:18 PM

P: 55

DE homework problem
I have y(t) = (tC)^2 when y>=0. I get the same thing when y<0 as well, by seperation of variables. I use tC 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
Feb1506, 06:19 AM

HW Helper
P: 1,024





#6
Feb1506, 06:35 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

"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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