Mixing Problems in Differential Equations 🧪

students, in this lesson you will learn how mixing problems turn real-life situations like salty water tanks, chemical solutions, and medicine concentration into differential equations. The big idea is simple: track how much of a substance enters and leaves a system over time. This topic fits into First-Order Equations II because the rate of change of the amount of substance is usually described by a first-order differential equation. By the end of this lesson, you should be able to explain the key terms, set up the equation, solve a typical problem, and connect the result to the broader study of first-order equations.

What is a Mixing Problem?

A mixing problem describes a system where a liquid contains some substance, and more liquid flows in and out. The substance might be salt, sugar, dye, pollution, or a medicine. The important question is usually: how much of the substance is in the tank at time $t$?

Here are the main pieces of a mixing problem:

$A(t)$ = amount of substance in the tank at time $t$
$V(t)$ = volume of liquid in the tank at time $t$
concentration = amount per volume, often written as $\frac{A(t)}{V(t)}$
inflow rate = how fast liquid enters, usually in liters per minute or gallons per minute
outflow rate = how fast liquid leaves, in the same units

The core principle is a balance equation:

$$\text{rate of change of amount} = \text{rate in} - \text{rate out}$$

That becomes

$$\frac{dA}{dt} = \text{inflow amount rate} - \text{outflow amount rate}.$$

This is why mixing problems belong in differential equations: the amount changes continuously, and the equation describes that change over time.

How the Model is Built

To solve a mixing problem, students, you usually follow the same steps. First, define the variable $A(t)$. Then figure out the volume $V(t)$, because concentration depends on it. Next, compute the amount of substance entering and leaving each minute.

If the inflow liquid has concentration $c_{\text{in}}$, then the amount of substance entering per unit time is

$$\text{rate in} = c_{\text{in}} \cdot r_{\text{in}},$$

where $r_{\text{in}}$ is the inflow rate.

If the tank is well-stirred, the concentration inside the tank is uniform, so the outflow concentration is

$$\frac{A(t)}{V(t)}.$$

Then the amount leaving per unit time is

$$\text{rate out} = \frac{A(t)}{V(t)} \cdot r_{\text{out}}.$$

So the general model is

$$\frac{dA}{dt} = c_{\text{in}}r_{\text{in}} - \frac{A(t)}{V(t)}r_{\text{out}}.$$

This is often a linear first-order differential equation, especially when $V(t)$ is constant or changes in a simple way.

Example 1: A Tank with Constant Volume

Suppose a tank holds $100$ liters of water. Salt water containing $2$ grams per liter flows in at $5$ liters per minute, and the mixed solution flows out at $5$ liters per minute. If the tank starts with $10$ grams of salt, find the amount of salt after time $t$.

Because inflow and outflow are equal, the volume stays constant at

$$V(t)=100.$$

Let $A(t)$ be the amount of salt in grams.

Step 1: Rate in

The inflow concentration is $2$ grams per liter, and the inflow rate is $5$ liters per minute, so

$$\text{rate in} = 2 \cdot 5 = 10 \text{ grams per minute}.$$

Step 2: Rate out

The concentration in the tank is

$$\frac{A(t)}{100}.$$

Since liquid leaves at $5$ liters per minute,

$$\text{rate out} = \frac{A(t)}{100}\cdot 5 = \frac{A(t)}{20}.$$

Step 3: Differential equation

So the equation is

$$\frac{dA}{dt} = 10 - \frac{A}{20}.$$

This is a first-order linear equation.

Step 4: Interpret the meaning

If $A(t)$ is very small, then the outflow term is small, so salt accumulates quickly. If $A(t)$ is large, then the outflow term increases, which slows the growth. This is a realistic feedback effect 🧪

The solution approaches a steady amount where

$$\frac{dA}{dt}=0.$$

That happens when

$$10 - \frac{A}{20}=0,$$

$$A=200.$$

That means the tank tends toward $200$ grams of salt over time.

Example 2: A Tank Whose Volume Changes

Some mixing problems are a little more interesting because the tank volume is not constant. Suppose a tank contains $50$ liters of solution. Brine enters at $3$ liters per minute with concentration $4$ grams per liter, and the mixed solution leaves at $2$ liters per minute. Let $A(t)$ be the amount of salt in grams.

Here the volume increases because more flows in than out:

$$V(t)=50+(3-2)t=50+t.$$

Now compute the rates.

Rate in

$$\text{rate in} = 4\cdot 3 = 12.$$

Rate out

The concentration in the tank is

$$\frac{A(t)}{50+t},$$

so the rate out is

$$\frac{A(t)}{50+t}\cdot 2 = \frac{2A(t)}{50+t}.$$

Differential equation

Thus,

$$\frac{dA}{dt}=12-\frac{2A}{50+t}.$$

This equation is still first-order, but now the coefficient depends on $t$. It is a linear equation that can be solved using an integrating factor, which connects mixing problems to the earlier part of First-Order Equations II.

students, this is a good example of how applied problems naturally lead to the methods from differential equations. The modeling step comes first, and the solution method comes second.

Why “Well-Stirred” Matters

A key assumption in mixing problems is that the tank is well-stirred. That means the substance is evenly distributed everywhere in the tank at each moment. Without that assumption, the concentration could be different in different parts of the tank, and the model would be much harder.

The well-stirred assumption lets us use one concentration value:

$$\frac{A(t)}{V(t)}.$$

This is what makes the outflow term easy to calculate. In real life, stirring helps make the model more accurate. In chemistry, water treatment, and industrial processing, this idea is very important because engineers often want to predict the concentration at future times.

Solving a Mixing Equation

Many mixing equations can be written in the standard linear form

$$\frac{dA}{dt}+p(t)A=q(t).$$

To solve this, we use an integrating factor

$$\mu(t)=e^{\int p(t)\,dt}.$$

Multiplying the whole equation by $\mu(t)$ turns the left side into a derivative:

$$\frac{d}{dt}\bigl(\mu(t)A(t)\bigr)=\mu(t)q(t).$$

Then we integrate both sides.

For example, if

$$\frac{dA}{dt}=12-\frac{2A}{50+t},$$

we rewrite it as

$$\frac{dA}{dt}+\frac{2}{50+t}A=12.$$

The integrating factor is

$$\mu(t)=e^{\int \frac{2}{50+t}\,dt}=e^{2\ln(50+t)}=(50+t)^2.$$

This step shows clearly how mixing problems connect to the broader technique of solving linear first-order equations.

Common Mistakes to Avoid

Mixing problems are very manageable when you keep the structure organized, but students often make the same mistakes. students, watch for these:

Confusing amount with concentration. The amount is $A(t)$; the concentration is $\frac{A(t)}{V(t)}$.
Using the wrong volume. If the volume changes, the denominator must be $V(t)$, not the starting volume.
Mixing up inflow and outflow units. Both rates must be compatible, usually volume per time.
Forgetting that the tank is well-stirred. Without that, the outflow concentration is not just a single value.
Writing only the equation and not defining variables. A good setup always starts with clear notation.

A neat way to check your work is to ask: does the equation make physical sense? For example, if no salt flows in, the amount should not increase forever unless there is already salt in the tank and no outflow.

Real-World Connections

Mixing models are used in many real situations. Water treatment plants use them to study pollutant levels in reservoirs. Food companies use them to control sugar or flavor concentrations. Hospitals can use similar reasoning for medication in the bloodstream, although more advanced models may be needed there.

These examples show why differential equations matter. They turn a real change process into a mathematical model that can be analyzed, predicted, and improved. Mixing problems are one of the clearest examples of this idea because the physical meaning of each term is easy to see.

Conclusion

Mixing problems are a major application of first-order differential equations. The central idea is the balance rule

$$\frac{dA}{dt}=\text{rate in}-\text{rate out}.$$

By defining the amount $A(t)$, finding the volume $V(t)$, and using concentration $\frac{A(t)}{V(t)}$, you can build a model for many real systems. When the equation is linear, integrating factors often help solve it. More importantly, mixing problems show how differential equations describe real-world change in a clear and useful way.

Study Notes

A mixing problem tracks how a substance enters and leaves a tank over time.
Let $A(t)$ be the amount of substance, usually in grams or kilograms.
The basic rule is $\frac{dA}{dt}=\text{rate in}-\text{rate out}$.
Concentration inside a well-stirred tank is $\frac{A(t)}{V(t)}$.
Rate in is usually $c_{\text{in}}r_{\text{in}}$.
Rate out is usually $\frac{A(t)}{V(t)}r_{\text{out}}$.
If inflow and outflow are equal, then the volume stays constant.
If inflow and outflow differ, then $V(t)$ changes with time.
Many mixing problems produce linear first-order differential equations.
Linear first-order equations can often be solved using an integrating factor $\mu(t)=e^{\int p(t)\,dt}$.
The well-stirred assumption is essential for using a single concentration value.
Mixing problems connect algebra, modeling, and calculus to real applications like chemistry, engineering, and medicine. 🙂