Thank you for your help,
I'll give you the answer for the growing case in i and ii
Note y = Ae^(kt)
y(0) = A
y = y(0)e^(kt) = y0 e^(kt)
if y(t1) = y1
Then y1 = y0 e^(kt1)
1/kln(y1/y0) = t1
Similarly for ii y(tn) = yn
yields 1/k ln(yn/y0) = tn
For your Lake problem You know your solution can't be correct
If P(t)=100,000e^(-t/1000) then P(0) =100,00
But if the concentration is initially .1% and there are 108 m^3
P(0) = .108m^3
It occurred to me you mean 10^8 m^3 and 10^5 m^3/day
otherwise if it were 108 and 105 you would have a mud puddle.
so dP/dt = pollution in -pollution out
dp/dt = 0 - %pollution*flow rate
dP/dt = -(P(t)/10^8) (10^5 m^2/day)
dP/dt = -P(t)/1000 P(0) = .001* 10^8 = 10^5
dp/P = -1/1000dt
ln(P) = -t/1000 +C
e^(ln(P) = e^(-t/1000+C) = e^C e^(-t/1000)
Since e^C is a constant we can just call it C
P = C*e^(-t/1000)
P(0) = 100,000 = C
P = 100,000 e^(-t/1000)
To find when it 1/10 original solve 10,000 = 100,000 e^(-t/1000)
1/10 = e^(-t/1000)
1/1000ln(10) = t