Solving a differential equation with initial conditions for s at t=10

[ex81]

Question: find the rate of change of (s) with respect to time(t), is inversely proportional to the square root of (s)

Write a differential equation for this statement.

Find the general solution to this equation

If initially (s)= 100, and after six seconds (s)= 144, what is the value of (s) be after 10th seconds?

Work so far:
Part one, ds=k/sqrt(s) dt

Part two, sqrt(s) ds = k dt
2/3(s)^3/2 = kt+c
(s)^3/2 = 3/2 kt +c
S=(3/2 kt +c)^2/3
So far the above is correct, and I know that these are true
T=0, s=100
T=6, s=144
T=10, s=?
The final answer is s=(6640/3)^2/3
I just don't know what to do to get the final answer...

 

[vela]

Use the information given to solve for k and c. For example, you're given s(0)=100. You also know s(0) = [3/2 k(0) + c]^2/3. Putting those two together, you can solve for c.

 

[ex81]

100^3/2 = c
Ergo 1000= c
144=(3/2 k 6 + 1000)^2/3
144^3/2 -1000= 9k
(144^3/2 -1000)/9=k
S=(3/18(144^3/2-1000)*10 +1000)^2/33
And I need a calculator to check that. Last time I did it the way, my answer did not match the final answer...

 

[vela]

$144^{3/2} = (\sqrt{144})^3 = 12^3 = 1728$

 

[ex81]

Thanks! Am a bit tired at the moment.
1728-1000= 728728
728*10= 7280
3/18*7280= 3640/3 + 3000/3
(6640/3)^3/2 which is the correct answer :-D
Thanks!