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10. Applications of First order Differential Equations

Dated: 10-11-2024

In order to translate a physical phenomena in terms of mathematics, we strive for a set of equations which describe the system adequately.
This set of equations is called Model for the phenomena.

Steps of Building a Model

  1. Clearly state the assumptions on which the model will be based. These assumptions should describe the relationships among quantities to be studied.
  2. Completely describe the parameters and variables to be used in the model.
  3. Use the assumptions from step 1 to derive the mathematical equations relating to parameters and variables from step 2.

Applications

  • Orthogonal Trajectories
  • Population dynamics
  • Radioactivity decay
  • Newton's law of cooling
  • Carbon dating
  • Chemical reactions

We know that solutions to first order linear differential equations1 can be given by implicit equations of the form

\[F(x, y, C) = 0\]

Where \(C\) is a family of curves.
Member curves can be obtained by fixing \(C\).

Similarly for an \(n^{\text{th}}\) order differential equation,

\[F(x, y, C_1, C_2, \ldots, C_n) = 0\]

Now the question is, can we represent an \(n^{\text{th}}\) order differential equation with \(n\) parameter family of curves as an \(n^{\text{th}}\) order differential equation with no parameters but still representing the family?
The answer in most cases is yes.

Some clues on how we can proceed.

  1. Differentiate with respect to \(x\) and get an equation involving \(x\), \(y\), \(\frac{dy}{dx}\) and \(C\).
  2. Using the original equation, we may be able to eliminate \(C\) from the new equation.
  3. Do some algebra and re-write the equation in explicit form
\[\frac{dy}{dx} = f(x, y)\]

Example

Let us say we have following families of curves.

\[ \begin{cases} y = mx \\ x^2 + y^2 = C^2 \end{cases} \]

The first family represents straight lines2 passing through the origin.
The second family represents circles centered at the origin.

Something like this
mth401_10_d_1.svg

It is clear that the lines2 are perpendicular to the tangent lines to the circles at the point of intersections.

Orthogonal Curves

Two curves \(C_1\) and \(C_2\) are called orthogonal if their tangent lines are perpendicular to each other at their point of intersection.

Orthogonal Trajectories

When all the curves of the family

\[\mathcal F_1 : G(x, y, c_1) = 0\]

orthogonally intersect all curves of another family

\[\mathcal F_2 : H(x, y, c_2) = 0\]

then each curve of a family is said to be orthogonal trajectory of another.

Occurrences

  • Fluid dynamics
  • Electricity and magnetism

Method of Finding Orthogonal Trajectory

Consider a family of curves \(\mathcal F\) and assume the associated differential equation is

\[\frac{dy}{dx} = f(x, y)\]

since \(\frac {dy}{dx}\) is giving slopes to the curves at \((x, y)\).
Therefore, the differential equation for the family of curves having orthogonal slopes will be

\[\frac{dy}{dx} = - \frac 1 {f(x, y)}\]

Summary

  • Consider a family of curves \(\mathcal F\) and the associated differential equations.
  • Re-write the differential equation in the form
\[\frac{dy}{dx}=f(x,y)\]
  • Write the differential equation associated with the orthogonal family.
\[\frac{dy}{dx}=-\frac{1}{f(x,y)}\]
  • Solve the equation and the solutions are the families of orthogonal curves.
  • A specific curve from the orthogonal family may be required, something like an initial value problem.3

Example

Find the orthogonal trajectory to the family of circles

\[x^2 + y^2 = C^2\]

Solution

\[2y\frac{dy}{dx}+2x=0\]
\[\frac{dy}{dx}=-\frac{x}{y}\]
\[\frac{dy}{dx}=-\frac{1}{- \frac x y}=\frac{y}{x}\]
\[u(x)=e^{-\int\frac{1}{x}dx}=\frac{1}{x}\]
\[y.u(x)=m\]
\[y=\frac{m}{u(x)}=mx\]

Population Dynamics

The easiest population dynamics model is exponential model.
This model is based on the following assumption

Assumption

The rate of change of the population is proportional to the existing population

If \(P(t)\) measures the population at time \(t\) then.

\[\frac{dP}{dt} = kP\]

Where \(k\) is the constant of proportionality.

\[\frac{d}{dt}[Pe^{-kt}]=0\]

Integrating4 both sides, we get

\[Pe^{-kt}=C\]
\[P=Ce^{kt}\]

if the initial population is

\[P(0) = P_0\]

so that

\[C = P_0\]

then

\[P(t)=P_{0}e^{kt}\]

Where \(k > 0\) for growth and \(k < 0\) for decay.

References

Read more about notations and symbols.

  1. Read more about first order linear differential equations. ↩

  2. Read more about lines. ↩↩

  3. Read more about initial value problem. ↩

  4. Read more about integration. ↩