Differential Equations - Modeling

Differential equations have been useful since its discovery because it extends to so many different fields or disciplines in an attempt to model behaviors. Disciplines which particularly love to use DEs are physics and engineering. This is mainly due to the fact that we can relate a function to its rate of change with respect to a certain variable makes it versatile to find that function, and thus in physics where we make observations based on our experiments/labs, it's easier to look at the rate of change (say velocity) to the actual function (displacement), rather than trying to get the function explicitly.

Talking about observations, it's noteworthy that many DEs are mainly based upon them. The steps toward getting a solution to the DE from modeling attempts are from making empirical observations, then going to construct the DE, and then finally solving the DE.

Differential equation modeling can be easy, but it can also very hard. We will start with baby steps, dealing with some easy problems and their explanations.

Let's start!

To model a differential equation, they will always give you information, such as rates of change, which can be expressed with differentials. Then when you express mathematically that information, you are able to continue and make some substitutions, or most commonly in easy questions, you can use the chain rule.

Let's look at an example:

Caleb has a cylindrical container of radius 2 cm and height 10 cm. A pipe fills the container with water at a rate of \(1 \text{ cm}^3\text{/s}\) and the depth increases at a rate of \(0.5\text{ cm/s}.\) What is the rate at which the volume of water increases with respect to its depth?

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We can start by formulating those derivatives that are given to us:

\[\frac{dV}{dt}=1\quad \text{ and } \quad \frac{dh}{dt}=\frac{1}{2}.\]

As a resut, we can find \(\frac{dV}{dh}\) by using the chain rule:

\[\frac{dt}{dh}\times\frac{dV}{dt}=2\times1\implies \frac{dV}{dh}=2.\]

And our final answer is 2. \( _\square \)

One of our founding fathers of calculus, Isaac Newton, derived Newton's law of cooling from a differential equation when he observed how the temperature of a liquid varies with the temperature surrounding it.

Newton's Law of Cooling

The rate of change of temperature of a body with instantaneous temperature \(T(t) \) is proportional to the difference between the body's temperature and that of the surrounding \(T_s: \)

\[\frac{d T(t)}{d t}=r\big(T(t)-T_{s}\big),\]

where \( r\) is the proportionality constant, which indicates how quick the body's temperature reverts to the surrounding's temperature.

Here the surrounding temperature is assumed constant. If it was not, the equation will be different and we will need to know what then the function \(T_s (t) \) is. This is beyond the scope of the subject. Nonetheless, we can solve this DE:

I will drop the \( (t) \). Remember that \( T \) is still a function of time.

If we may let \[ y = T - T_{s},\]

then because \( T_s \) is constant, i.e. \( \frac{dT_{s}}{dt} = 0, \)

\[ \frac{dy}{dt} = \frac{d(T - T_{s})}{dt} = \frac{dT}{dt} - \frac{dT_{s}}{dt} = \frac{dT}{dt} = r(T - T_{s}) = ry \]

or basically

\[\frac{dy}{dt} = ry. \]

Separating the differentials,

\[\frac{dy}{y} = r dt. \]

Integrate both sides:

\[\begin{align} \int \frac{dy}{y} &= \int r dt \\ \ln y &= \ln (T - T_{s})\\ &= rt + C. \end{align} \]

To find \(C,\) we let \( t=0 \) to get that

\[ \ln \big(T(0) - T_{s})\big) = C. \]

Hence,

\[\begin{align} T - T_{s} &= e^{rt + C} \\ &= e^{C} e^{rt} \\ &= \big(T(0) - T_{s}\big)e^{rt} \\ \Rightarrow T &= T_s + \big(T(0) - T_{s}\big)e^{rt}. \end{align}\]

Therefore, we got a function for how the temperature of an object changes over time when it has a temperature difference to the surrounding temperature.