Thread: Diff Eq - Existence/uniqueness theorem

1. Diff Eq - Existence/uniqueness theorem

So the question is:
Using the existence and uniqueness theorem, show that the differential equation dy/dt = sqrt(y-1), with initial condition that y(0)=2, has a unique solution. Furthermore, solve it.

I verified that it both existed and was unique, and I also got a solution but I'm not quite sure if it's right...

I separated it into 1/sqrt(y-1) * dy = 1 dt
integrated: 2 sqrt(y-1) = t + C
divide by 2: sqrt(y-1) = t/2 + C

now this is where I'm not sure if I messed up or not. Silly, since it's simple algebra, but when i remove the square root from the y-1, would the right side turn into (t^2)/4 + C, or (t^2)/4 + Ct + C? I chose the former and got a final answer of y(t) = (t^2)/4 + 2
I'm totally braindead right now, haha, but should I have done it the second way?

2. Originally Posted by mistykz
So the question is:
Using the existence and uniqueness theorem, show that the differential equation dy/dt = sqrt(y-1), with initial condition that y(0)=2, has a unique solution. Furthermore, solve it.

I verified that it both existed and was unique, and I also got a solution but I'm not quite sure if it's right...

I separated it into 1/sqrt(y-1) * dy = 1 dt
integrated: 2 sqrt(y-1) = t + C (**)
divide by 2: sqrt(y-1) = t/2 + C

now this is where I'm not sure if I messed up or not. Silly, since it's simple algebra, but when i remove the square root from the y-1, would the right side turn into (t^2)/4 + C, or (t^2)/4 + Ct + C? I chose the former and got a final answer of y(t) = (t^2)/4 + 2
I'm totally braindead right now, haha, but should I have done it the second way?
If

$\displaystyle \sqrt{y-1} = \frac{t}{2} + c$ then $\displaystyle y - 1 = \left( \frac{t}{2} + c\right)^2$

but I would personally bring in the IC at the step (**)