Question

A 400 L tank is filled with pure water. A copper sulfate solution with a concentration of 20 g/L flows into the tank at a rate of 4 L/min. The thoroughly mixed solution is drained from the tank at a rate of 4 L/min. a. Write a differential equation (initial value problem) for the mass of the copper sulfate. b. Solve the differential equation

Answer 1

(a) Let $C(t)$ denote the amount (in grams) of copper (II) sulfate (CuSO₄) in the tank at time $t$ minutes. The tank contains only pure water at the start, so we have initial value $\boxed{C(0)=0}$.

CuSO₄ flows into the tank at a rate

$\left(20\dfrac{\rm g}{\rm L}\right) \left(4\dfrac{\rm L}{\rm min}\right) = 80 \dfrac{\rm g}{\rm min}$

and flows out at a rate

$\left(\dfrac{C(t)\,\rm g}{400\,\mathrm L + \left(4\frac{\rm L}{\rm min} - 4\frac{\rm L}{\rm min}\right) t}\right) \left(4\dfrac{\rm L}{\rm min}\right) = \dfrac{C(t)}{100} \dfrac{\rm g}{\rm min}$

and hence the net rate of change in the amount of CuSO₄ in the tank is governed by the differential equation

$\boxed{\dfrac{dC}{dt} = 80 - \dfrac C{100}}$

(b) This ODE is linear with constant coefficients and separable, so we have a few choices in how we can solve it. I'll use the typical integrating factor method for solving linear ODEs.

$\dfrac{dC}{dt} + \dfrac C{100} = 80$

The integrating factor is

$\mu = \exp\left(\displaystyle \int \frac{dt}{100}\right) = e^{t/100}$

Distributing $\mu$ on both sides gives

$e^{t/100} \dfrac{dC}{dt} + \dfrac1{100} e^{t/100} C = 80 e^{t/100}$

and the left side is now the derivative of a product,

$\dfrac d{dt} \left[e^{t/100} C\right] = 80 e^{t/100}$

Integrate both sides. By the fundamental theorem of calculus,

$e^{t/100} C = e^{t/100}C\bigg|_{t=0} + \displaystyle \int_0^t 80 e^{u/100}\, du$

The first term on the right vanishes since $C(0)=0$. Then

$e^{t/100} C = 8000 \left(e^{t/100} - 1\right)$

$\implies \boxed{C(t) = 8000 - 8000 e^{-t/100}}$