Application of Differential Equations: Definition, Types, Examples

Applications of Differential Equations: A differential equation, also abbreviated as D.E., is an equation for the unknown functions of one or more variables. It relates the values of the function and its derivatives. Differential equations have applications in various fields of Science like Physics (dynamics, thermodynamics, heat, fluid mechanics, and electromagnetism), Chemistry (rate of chemical reactions, physical chemistry, and radioactive decay), Biology (growth rates of bacteria, plants and other organisms) and Economics (economic growth rate, and population growth rate). Having said that, almost all modern scientific investigations involve differential equations.

Differential Equation: Overview

The equation that involves independent variables, dependent variables and their derivatives is called a differential equation.

There are two types of differential equations:

• Ordinary Differential Equations
• Partial Differential Equations

Applications of Differential Equations in Real Life

The applications of differential equations in real life are as follows:

In Physics:

• Study the movement of an object like a pendulum
• Study the movement of electricity
• To represent thermodynamics concepts

In Medicine:

• Graphical representations of the development of diseases

In Mathematics:

Describe mathematical models such as:

• population explosion
• radioactive decay

Geometrical applications:

To find the:

• slope of a tangent
• equation of tangent and normal
• length of tangent and normal
• length of sub-tangent and sub-normal

Physical application:

We can calculate

• velocity
• acceleration

Applications of First-order Differential Equations

The applications of the First-order differential equations are as follows:

What is an ODE?

An ordinary differential equation, or ODE, is a differential equation in which the dependent variable is a function of the independent variable. The highest order derivative in the differential equation is called the order of the differential equation.

First-Order ODE

The first-order differential equation is given by

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

Where,

\(x\) is the independent variable

\(y\) is the dependent variable.

It has only the first-order derivative \(\frac{{dy}}{{dx}}\). Hence, the order is \(1\).

First-order differential equations have a wide range of applications.

Growth and Decay

Let \(N(t)\) denote the amount of substance (or population) that is growing or decaying. If we assume that the time rate of change of this amount of substance, \(\frac{{dN}}{{dt}}\), is proportional to the amount of substance present, then

\(\frac{{dN}}{{dt}} = kN\), or \(\frac{{dN}}{{dt}} – kN = 0\)

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

Here, we assume that \(N(t)\) is a differentiable, continuous function of time.

Newton’s Cooling Law

Newton’s law of cooling and heating, states that the rate of change of the temperature in the body, \(\frac{{dT}}{{dt}}\), is proportional to the temperature difference between the body and its medium. Newton’s law of cooling can be formulated as

\(\frac{{dT}}{{dt}} = – k\left( {T – {T_m}} \right)\)

\( \Rightarrow \frac{{dT}}{{dt}} + kT = k{T_m}\)

Where \(k\) is a positive constant of proportionality.

Applications of Second-order Differential Equations

The applications of second-order differential equations are as follows:

Second-Order ODE

The second-order differential equation is given by

\({y^{\prime \prime }} + p(x){y^\prime } + q(x)y = f(x)\)

Where,

\(x\) is the independent variable

\(y\) is the dependent variable.

\(p\left( x \right)\) and \(q\left( x \right)\) are either constant or function of \(x\).

The highest order derivative is \(\frac{{{d^2}y}}{{d{x^2}}}\). Hence, the order is \(2\).

Note:

1. If \(f(x) = 0\), then the equation becomes a homogeneous second order differential equation.
2. If \(f(x) \ne 0\), then the equation becomes a non-homogeneous second order differential equation.

Second-order differential equations have a wide range of applications.

Systems of Electrical Circuits

Electrical systems, also called circuits or networks, are designed as combinations of three components: resistor \(\left( {\rm{R}} \right)\), capacitor \(\left( {\rm{C}} \right)\), and inductor \(\left( {\rm{L}} \right)\). They are defined by resistance, capacitance, and inductance and is generally considered lumped-parameter properties.

The second-order differential equation has derivatives equal to the number of elements storing energy. This differential equation is considered an ordinary differential equation.

Simple Harmonic Motion

The differential equation for the simple harmonic function is given by

 \(\frac{{{d^2}x}}{{d{t^2}}} = – {\omega ^2}x\), where \(\omega \) is the angular velocity of the particle and \(T = \frac{{2\pi }}{\omega }\) is the period of motion.

From this, we can conclude that for the larger mass, the period is longer, and for the stronger spring, the period is shorter.

Applications of Partial Differential Equations

The applications of partial differential equations are as follows:

What is PDE?

A Partial differential equation (or PDE) relates the partial derivatives of an unknown multivariable function. Such a multivariable function can consist of several dependent and independent variables.

Examples:

Laplace Equation: \({\Delta ^2}\phi = \frac{{{\partial ^2}\phi }}{{{\partial ^2}x}} + \frac{{{\partial ^2}\phi }}{{{\partial ^2}y}} = 0\)

Heat Conduction Equation: \(\frac{{\partial T}}{{\partial t}} = C\frac{{{\partial ^2}T}}{{\partial {x^2}}}\)

In physical problems:

• To calculate the boundary conditions
• To solve boundary value problems using the method of separation of variables

Several problems in Engineering give rise to some well-known partial differential equations. Few of them are listed below.

1. Wave Equation

\(\frac{{{\partial ^2}T}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}y}}{{\partial {x^2}}}\)

2. One dimensional heat flow equation

\(\frac{{\partial u}}{{\partial t}} = {c^2}\frac{{{\partial ^2}T}}{{\partial {x^2}}}\)

3. Two dimensional heat flow equation which is steady state becomes the two dimensional Laplace’s equation

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0\)

4. Laplace’s equation in three dimensions

\({\Delta ^2}\phi = \frac{{{\partial ^2}\phi }}{{{\partial ^2}x}} + \frac{{{\partial ^2}\phi }}{{{\partial ^2}y}} + \frac{{{\partial ^2}\phi }}{{{\partial ^2}z}} = 0\)

Solved Examples

Q.1. The population of a country is known to increase at a rate proportional to the number of people presently living there. If after two years the population has doubled, and after three years the population is \(20,000\), estimate the number of people currently living in the country.
Ans:
Let \(N\) denote the number of people living in the country at any time \(t\), and let \({N_0}\) denote the number of people initially living in the country.
\(\frac{{dN}}{{dt}}\), the time rate of change of population is proportional to the present population.
Then \(\frac{{dN}}{{dt}} = kN\), or \(\frac{{dN}}{{dt}} – kN = 0\), where \(k\) is the constant of proportionality.
\(\frac{{dN}}{{dt}} – kN = 0\) which has the solution \(N = c{e^{kt}}…….(i)\)
At \(t = 0,\,N = {N_0}\)
Hence, it follows from \((i)\) that \(N = c{e^{k0}}\)
\( \Rightarrow {N_0} = c{e^{k0}}\)
\(\therefore \,{N_0} = c\)
Thus, \(N = {N_0}{e^{kt}}\,………(ii)\)
At \(t = 2,\,N = 2{N_0}\) [After two years the population has doubled]
Substituting these values into \((ii)\),
We have \(2{N_0} = {N_0}{e^{kt}}\) from which \(k = \frac{1}{2}\ln 2\)
Substituting these values into \((i)\) gives
\(N = {N_0}{e^{\frac{t}{2}(\ln 2)}}\,…….(iii)\)
At \(t = 3,\,N = 20000\).
Substituting these values into \((iii)\), we obtain
\(20000 = {N_0}{e^{\frac{3}{2}(\ln 2)}}\)
\({N_0} = \frac{{20000}}{{2\sqrt 2 }} \approx 7071\)
Hence, \(7071\) people initially living in the country.

Q.2. A metal bar at a temperature of \({100^{\rm{o}}}F\) is placed in a room at a constant temperature of \({0^{\rm{o}}}F\). If, after \(20\) minutes, the temperature is \({50^{\rm{o}}}F\), find the time to reach a temperature of \({25^{\rm{o}}}F\).
Ans:
Newton’s law of cooling is \(\frac{{dT}}{{dt}} = – k\left( {T – {T_m}} \right)\)
\( \Rightarrow \frac{{dT}}{{dt}} + kT = k{T_m}\)
\( \Rightarrow \frac{{dT}}{{dt}} + kT = 0\,\,\left( {\therefore \,{T_m} = 0} \right)\)
Which has the solution \(T = c{e^{ – kt}}\,…….(i)\)
Since \(T = 100\) at \(t = 0\)
\(\therefore \,100 = c{e^{ – k0}}\) or \(100 = c\)
Substituting these values into \((i)\) we obtain
\(T = 100{e^{ – kt}}\,……..(ii)\)
At \(t = 20\), we are given that \(T = 50\); hence, from \((ii)\),
\(50 = 100{e^{ – kt}}\) from which \(k = – \frac{1}{{20}}\ln \frac{{50}}{{100}}\)
Substituting this value into \((ii)\), we obtain the temperature of the bar at any time \(t\) as \(T = 100{e^{\left( {\frac{1}{{20}}\ln \frac{1}{2}} \right)t}}\,………(iii)\)
When \(T = 25\)
\(25 = 100{e^{\left( {\frac{1}{{20}}\ln \frac{1}{2}} \right)t}}\)
\( \Rightarrow t = 39.6\) minutes
Hence, the bar will take \(39.6\) minutes to reach a temperature of \({25^{\rm{o}}}F\).

Q.3. A tank initially holds \(100\,l\) of a brine solution containing \(20\,lb\) of salt. At \(t = 0\), fresh water is poured into the tank at the rate of \({\rm{5 lit}}{\rm{./min}}\), while the well stirred mixture leaves the tank at the same rate. Find amount of salt in the tank at any time \(t\).
Ans:
Here, \({V_0} = 100,\,a = 20,\,b = 0\), and \(e = f = 5\),
Now, from equation \(\frac{{dQ}}{{dt}} + f\left( {\frac{Q}{{\left( {{V_0} + et – ft} \right)}}} \right) = be\), we get
\(\frac{{dQ}}{{dt}} + \left( {\frac{1}{{20}}} \right)Q = 0\)
The solution of this linear equation is \(Q = c{e^{\frac{{ – t}}{{20}}}}\,………(i)\)
At \(t = 0\) we are given that \(Q = a = 20\)
Substituting these values into \((i)\), we find that \(c = 20\) so that \((i)\) can be rewritten as
\(Q = 20{e^{\frac{{ – t}}{{20}}}}\)
Note that as \(t \to \infty ,\,Q \to 0\) as it should since only freshwater is added.

Q.4. Find the equation of the curve for which the Cartesian subtangent varies as the reciprocal of the square of the abscissa.
Ans:
Let \(P(x,\,y)\) be any point on the curve, according to the question
Subtangent \( \propto \frac{1}{{{x^2}}}\) or \(y\frac{{dx}}{{dy}} = \frac{k}{{{x^2}}}\)
Where \(k\) is constant of proportionality or \(\frac{{kdy}}{y} = {x^2}dx\)
Integrating, we get \(k\ln y = \frac{{{x^3}}}{3} + \ln c\)
Or \(\ln \frac{{{y^k}}}{c} = \frac{{{x^3}}}{3}\)
\({y^k} = {c^{\frac{{{x^3}}}{3}}}\)  which is the required equation.

Q.5. Solve the equation \(\frac{{\partial u}}{{\partial t}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) with boundary conditions \(u(x,\,0) = 3\sin \,n\pi x,\,u(0,\,t) = 0\) and \(u(1,\,t) = 0\) where \(0 < x < 1,\,t > 0\).
Ans: The solution of differential equation \(\frac{{\partial u}}{{\partial t}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\,……..(i)\)
is \(u(x,\,t) = \left( {{c_1}\,\cos \,px + {c_2}\,\sin \,px} \right){e^{ – {p^2}t}}\,……..(ii)\)
When \(x = 0,\,u(0,\,t) = {c_1}{e^{ – {p^2}t}} = 0\) i.e., \({c_1} = 0\).
Therefore \((ii)\) becomes \(u(x,\,t) = {c_2}\,\sin \,px{e^{ – {p^2}t}}\,……….(iii)\)
When \(x = 1,\,u(1,\,t) = {c_2}\,\sin \,p \cdot {e^{ – {p^2}t}} = 0\) or \(\sin \,p = 0\)
i.e., \(p = n\pi \).
Therefore, \((iii)\) reduces to \(u(x,\,t) = {b_n}{e^{ – {{(n\pi )}^2}t}}\sin \,n\pi x\) where \({b_n} = {c_2}\)
Thus the general solution of \((i)\) is \(u(x,\,t) = \sum {{b_n}} {e^{ – {{(n\pi )}^2}t}}\sin \,n\pi x\,……….(iv)\)
When \(t = 0,\,3\,\sin \,n\pi x = u(0,\,t) = \sum\limits_{n = 1}^\infty {{b_n}} {e^{ – {{(n\pi )}^2}t}}\sin \,n\pi x\)
Comparing both sides, \({b_n} = 3\)
Hence from \((iv)\), the desired solution is
\(u(x,\,t) = 3\sum\limits_{n = 1}^\infty {{b_n}} {e^{ – {{(n\pi )}^2}t}}\sin \,n\pi x\)

Learn About Methods of Solving Differential Equations

Summary

An equation that involves independent variables, dependent variables and their differentials is called a differential equation. The first-order differential equation is defined by an equation \(\frac{{dy}}{{dx}} = f(x,\,y)\),  here \(x\) and \(y\) are independent and dependent variables respectively. The main applications of first-order differential equations are growth and decay, Newton’s cooling law, dilution problems.

Partial differential equations relate to the different partial derivatives of an unknown multivariable function. Several problems in engineering give rise to partial differential equations like wave equations and the one-dimensional heat flow equation. Similarly, the applications of second-order DE are simple harmonic motion and systems of electrical circuits.

Frequently Asked Questions (FAQs)

Q.1. What is a differential equation and its application?
Ans: An equation that has independent variables, dependent variables and their differentials is called a differential equation. They are used to calculate the movement of an item like a pendulum, movement of electricity and represent thermodynamics concepts. Graphical representations of the development of diseases are another common way to use differential equations in medical uses.

Q.2. What are the applications of differential equations in engineering?
Ans: It has vast applications in fields such as engineering, medical science, economics, chemistry etc. Application of differential equations in engineering are modelling of the variation of a physical quantity, such as pressure, temperature, velocity, displacement, strain, stress, voltage, current, or concentration of a pollutant, with the change of time or location, or both would result in differential equations.

Q.3. What are the applications of differential equations?
Ans: Differential equations have many applications, such as geometrical application, physical application. In geometrical applications, we can find the slope of a tangent, equation of tangent and normal, length of tangent and normal, and length of sub-tangent and sub-normal.

Q.4. How many types of differential equations are there?
Ans: There are 6 types of differential equations. They are as follows:

1. Ordinary Differential Equations
2. Partial Differential Equations
3. Linear Differential Equations
4. Non-linear Differential Equations
5. Homogeneous Differential Equations
6. Non-homogeneous Differential Equations

Q.5. What are the applications of differentiation in economics?
Ans: The application of differential equations in economics is optimizing economic functions. For example, the use of the derivatives is helpful to compute the level of output at which the total revenue is the highest, the profit is the highest and (or) the lowest, marginal costs and average costs are the smallest.