Modeling with First-Order ODEs
Differential Equations · Section 2.3 · generated practice set
Mixing tanks, Newton's law of cooling and falling bodies with drag — rate in minus rate out before any calculus.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
A tank holds 400 L of well-stirred solution containing 30 g of salt. Brine of concentration 1 g/L flows in at 10 L/min, and the mixture flows out at the same rate. Let $Q(t)$ be the grams of salt in the tank at time $t$ (minutes).
Solution
- Rate in $= c\,r = 1\cdot10 = 10$ g/min. Rate out $= r\cdot\frac{Q}{V} = \frac{10}{400}Q$.
- So $Q' = 10 - 0.025Q$, a first-order linear equation with $Q(0) = 30$.
- Equilibrium (limiting) value: set $Q' = 0 \Rightarrow Q = \frac{10}{0.025} = cV = 400$ g.
- General solution: $Q(t) = 400 + \left(30 - 400\right)e^{-0.025t} = 400 - 370e^{-0.025t}$.
- At $t = 20$: $Q \approx 175.5837$ g.
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Method
Modeling problems are graded on the setup as much as the algebra. Write the balance law before you write any calculus.
$$\frac{d(\text{amount})}{dt} = \text{rate in} - \text{rate out}$$
Mixing tanks
Rate in $= (\text{inflow concentration}) \times (\text{inflow rate})$. Rate out $= (\text{outflow rate}) \times \dfrac{Q}{V(t)}$ — the concentration leaving is the tank's concentration. If inflow and outflow rates match, $V$ is constant and $$Q' = rc - \frac{r}{V}Q,\qquad Q(\infty) = cV.$$ If they differ, $V(t) = V_0 + (r_{in} - r_{out})t$ and the equation is still linear, just with a $t$ in the coefficient.
Newton's law of cooling
$T' = -k(T - T_{amb})$. Substituting $u = T - T_{amb}$ gives $u' = -ku$, so the difference decays exponentially. Two data points determine $k$.
Falling body with linear drag
$mv' = mg - \gamma v$. Terminal velocity $= mg/\gamma$, obtained by setting $v' = 0$ — no solving required.
Definitions and theorems
Worked example
A 200 L tank starts with 30 g of salt. Brine at 2 g/L enters at 4 L/min; the mixture leaves at 4 L/min. Find $Q(t)$ and the limiting amount.
- Rate in $= 2 \times 4 = 8$ g/min.
- Rate out $= 4 \cdot \frac{Q}{200} = 0.02Q$ g/min (volume is constant since the rates match).
- $Q' = 8 - 0.02Q$, $Q(0) = 30$.
- Equilibrium: $Q' = 0 \Rightarrow Q = 400$ g. (Check: $cV = 2 \times 200 = 400$.)
- $Q(t) = 400 + (30 - 400)e^{-0.02t} = 400 - 370e^{-0.02t}$, and $Q \to 400$ g.
Common mistakes
- Using the inflow concentration for the outflow. What leaves has the tank's current concentration $Q/V$, which changes.
- Assuming $V$ is constant. Check the two flow rates before writing $Q/V$.
- Unit drift. Minutes vs hours, liters vs gallons, grams vs kg — write units on every rate.
- Solving the whole ODE for a limiting value. Set $y' = 0$ instead; it is one line.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.