The question is as follows:
I am confused because of intial and after 30 mins concentration dont know how to split that. Still here is what i tried:
concentration initially = $2\times 2 = 4000g/min$
after 30 min = $2\times 1 = 2000g/min$
Inflow rate : $di/dt = 2000$ after 30 mins
outflow rate : $do/dt = 2 l/min$
Now not sure how to implement $y(t)$ here
after 30 mins will it be ? $$4000-\frac{y}{10}=y'$$ ?
Best Answer
You seem to be trying to put 30 minutes directly as a term of the differential equation somewhere. That's not the case. The differential equation is what it is, until the influx changes.
In a time $dt$, you lose mass due to current concentration $c(t)$ flowing out, and add mass due to inflow concentration $c_0$ flowing in.
$$dm = -c(t) dV + c_0 dV$$ Here we took into account, that the inflow and outflow volume $dV$ is the same (you just displace the old water with new more salty water). Dividing by $dt$ we get the volume flux $\Phi_V$ on the right side. $$\frac{dm}{dt} = (c_0-c(t))\Phi_V$$ Of course, concentration is $c=m/V_0$ ($V_0$ being the fixed whole container volume), so if we divide by $V_0$ we get the differential equation for $c$. $$\frac{dc}{dt} = (c_0-c)\frac{\Phi_V}{V_0}$$ If we rename concentration to $y$ to match your notation, and rearrange, you get $$\boxed{y'+\frac{\Phi_V}{V_0}y = c_0\frac{\Phi_V}{V_0}}$$
The coefficients are: $$\frac{\Phi_V}{V_0} = \frac{2\,\rm l/min}{20\,\rm l}=0.1\,\rm min^{-1}$$ The right side changes, first half hour it equals: $$\frac{c_0 \phi_V}{V_0} = 2{\,\rm kg/l}\times 0.1\,\rm min^{-1}=0.2 {\,\rm kg\,l^{-1}\,min^{-1}}$$ and then it becomes (half of that before) $$0.1 {\,\rm kg\,l^{-1}\,min^{-1}}$$ p.s. don't ignore units, they tell you immediately if your differential equation makes sense or not.
The boxed equation is trivially solved by an exponential approach to the final value (which is obvious without calculation -- waiting for long enough, the concentration will match the inflow concentration). Solution: $$y(t)=c_0 + (y(0)-c_0)e^{-t \Phi_V/V_0}$$ First, you have pure water, so $y(0)=0$ and you use the $c_0$ from the first half hour. For the second part, your initial $y(0)$ is whatever you got from the first half hour, and $c_0$ is now different, but the general solution is the same.