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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

User Harism
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(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(\rm g)/(\rm L)\right) \left(4(\rm L)/(\rm min)\right) = 80 (\rm g)/(\rm min)

and flows out at a rate


\left((C(t)\,\rm g)/(400\,\mathrm L + \left(4(\rm L)/(\rm min) - 4(\rm L)/(\rm min)\right) t)\right) \left(4(\rm L)/(\rm min)\right) = (C(t))/(100) (\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{(dC)/(dt) = 80 - \frac 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.


(dC)/(dt) + \frac C{100} = 80

The integrating factor is


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

Distributing
\mu on both sides gives


e^(t/100) (dC)/(dt) + \frac1{100} e^(t/100) C = 80 e^(t/100)

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


\frac 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)}

User Vtokmak
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