Hello ZZZZZzzz,
We have an initial amount of pollution in the lake, and we have pollution flowing in and pollution flowing out. So we may model the amount of pollution $P$ in the lake at time $t$ with the initial value problem:
$$\frac{dP}{dt}=0.2\cdot2\cdot10^4-\frac{P(t)}{10^6}\cdot2\cdot10^4$$ where $$P_0=2\cdot10^6$$
The ODE may be simplified to:
$$\frac{dP}{dt}=4000-\frac{P(t)}{50}$$
Writing this ODE in standard linear form, we have:
$$\frac{dP}{dt}+\frac{1}{50}P(t)=4000$$
Multiplying through by an integrating factor of $$e^{\frac{t}{50}}$$, we obtain:
$$e^{\frac{t}{50}}\frac{dP}{dt}+\frac{1}{50}e^{\frac{t}{50}}P(t)=4000e^{\frac{t}{50}}$$
Observing the left side is the differentiation of a product, we may write:
$$\frac{d}{dt}\left(e^{\frac{t}{50}}P(t) \right)=4000e^{\frac{t}{50}}$$
Integrating, there results:
$$e^{\frac{t}{50}}P(t)=200000e^{\frac{t}{50}}+C$$
Solving for $P(t)$, we have:
$$P(t)=200000+Ce^{-\frac{t}{50}}$$
To determine the parameter $C$, we may use the initial value:
$$P(0)=200000+C=2000000\implies C=1800000$$
Hence:
$$P(t)=200000+1800000e^{-\frac{t}{50}}=200000\left(1+9e^{-\frac{t}{50}} \right)$$
Now, to determine when the pollution concentration is down to 1 gram per million cubic meters, we set $P(t)=1000000$ and solve for $t$:
$$1000000=200000\left(1+9e^{-\frac{t}{50}} \right)$$
$$5=1+9e^{-\frac{t}{50}}$$
$$\frac{9}{4}=e^{\frac{t}{50}}$$
Convert from exponential to logarithmic form:
$$\frac{t}{50}=\ln\left(\frac{9}{4} \right)$$
$$t=50\ln\left(\frac{9}{4} \right)\approx40.5465108108164$$
Thus, we find it will take about 40.5 days for the pollution concentration to reach the desired level.
