Back to the Basics: Differential Equations

Calculus and differential equations have immense utility in any field of engineering. Any quality engineer should take it upon themselves to occasionally review their academic work so that they maintain their knowledge base. I will try to skip over the mathematical theory and get right to solving certain types of differential equations since as engineers, we typically only need the concrete concepts.

Here is a practical example of a first order mixing solution problem. We are mixing a salt in a cylindrical tank. We have several variables: the concentration of salt going into and leaving the tank Cin and Cout respectively, the rate of water going into and leaving the tank, rin and rout respectively, the amount of water in the tank S(t), and the amount of salt in the water at time t, Q(t). Assuming that water leaving the tank is the same concentration as the water in the tank, Cout can be simplified by Q(t)/S(t). How do we find the amount of salt in the tank at time t?

Here are some initial conditions to help get us started. At time t=0 the amount of water in the tank is 500 liters, there are 100 grams of salt in the tank, the rate of water going into and out of the tank is 5 liters/min and the concentration going in is 2 grams/min. We know now that the level of water is not going to change since the flow rates are equal. Let’s look at our equation to determine salt levels over time.

Q'(t) = rin*Cin – rout*Q(t)/S(t)

Now we replace values that we already know and get:

Q'(t) = 5 * 2 – 5 * Q(t) / 500 or Q'(t) = 10 – Q(t) /100

Group like variables on opposite sides and you should get:

Q'(t) + Q(t) / 100 = 10

At this point we use a substitution method to simplify the equation. This step is especially important if you have length functions and derivatives. We simplify the function such that the derivative has no coefficient as is already done. This is the form we are aiming for:

μ(t)*Q'(t) +   μ(t)*p(t)*Q(t)/100 = g(t)

Here p(t) is 1/100 and g(t) is 10. We are going to solve for  μ(t). μ must satisfy the equation μ'(t)=p(t)*μ(t). The simplified form is below:

μ'(t)/ μ(t) = 10

Integrate both sides. Solve for  μ(t) by eliminating with the exponential function. You should get.

μ(t) = et/100

This allows us to integrate the right side to get μ(t)*Q(t) the left side is ∫g(t)*μ(t)dt. Note the dt in that equation. dt in this case is 100. This is what we are left with:

Q(t)=et/100 * ∫10*100*et/100

The result of this integral will produce a constant too, which will require the initial condition so it can be solved. Here is the solved form:

Q(t) = 1000+C*e-t/100

Since we know at time t = 0 Q is 100, we can then solve for C and then we can use this equation to determine salt levels over time. I’m going to leave that to you do to do for practice if you so choose. I realize my math is a bit sloppy, let me know if you have any improvements. Thanks for reading!


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