Basics of Differential Equations: Have you heard the saying, “Change is the only constant”? Anything in this world that changes can be represented mathematically as a differential equation. Examples of real-life applications are the increase in the population of a country, continuous compounding of interest, and the rate of decay of radioactive materials. Although there are several applications of differential equations, they are hard to use. Hence, they are often used as simple equations without the differential terms. We can then use these to make calculations, predict solutions and represent them on graphs. This article explains ordinary differential equations, their types, and solutions.

Types of Differential Equations

There are different types of differential equations:
1. Ordinary differential equations
2. Partial differential equations
3. Linear differential equations
4. Nonlinear differential equations
5. Homogenous differential equations
6. Nonhomogenous differential equations

What is an Ordinary Differential Equation?

In general, a differential equation is that which has a function and its derivative. An ordinary differential equation often referred to as ODE, is an equation that involves a function in one variable and at least one of its derivatives. The term ordinary is included to differentiate these from partial differential equations.

Examples of ordinary differential equations:
1. \(x + x’ = 0\)
2. \(\frac{{dx}}{{dt}} = 5x – 3\)
3. \(\frac{{{d^2}y}}{{d{x^2}}} = 4{y^3}{x^2}\)

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Simplest Differential Equation

The simplest form of an ordinary differential equation is given by:
\(\frac{{dy}}{{dx}} = f\left( x \right)\)
Where,
\(f \to \) given function
If \(f\left( x \right) = 3{x^2},\) then it means that the derivative of the dependent variable \(y\left( x \right)\) is \(3{x^2}.\)

Order of an Ordinary Differential Equation (ODE)

The differential equations are both defined and classified by their order. The order of an ordinary differential equation is the highest derivative that appears in the equation.

Type of ODE	Example	Order of the ODE
First-order differential equation	\(\frac{{dy}}{{dx}} + \cos \,x = 0\)	\(1\)
Second-order differential equation	\(y” – 25y = 0\)	\(2\)
Third-order differential equation	\(z’’’ – 9z’’ + 15z’ + 25z = 0\)	\(3\)
Fourth-order differential equation	\(\frac{{{d^4}y}}{{d{t^4}}} + x = 0\)	\(4\)

Note that the order of an ordinary differential equation is always a positive number.

Degree of an Ordinary Differential Equation (ODE)

The degree of an ordinary differential equation is the exponent of the highest order derivative in that equation. Similar to the order, the degree of a differential equation is also a positive number.
Example:
\(\frac{{{d^3}y}}{{d{x^3}}} + {x^2}\frac{{dy}}{{dx}} = 0\)
Here,
1. The highest derivate is \(3,\) and hence, the order of the differential equation is \(3.\)
2. The exponent of the variable in the highest order derivative is \(1,\) and hence, the degree of the differential equation is \(1.\)
Note: While finding the degree of an ordinary differential equation, it is valid only if the function is a polynomial. If the function is trigonometric, logarithmic, or exponential, then the degree of that differential equation stands undefined.
Example:
\(\frac{{dy}}{{dx}} + {\rm{cos}}\,x = 0\)
Here, the order of the differential equation is \(1,\) but the degree is not defined.

Solution of an Ordinary Differential Equation (ODE)

A solution of an ODE is a function that satisfies the given differential equation. It is written as the value of the dependent variable in terms of the independent variable. There are two types of solutions to an ordinary differential equation.
1. General solution
2. Particular solution

General Solution of an Ordinary Differential Equation (ODE)

Consider the differential equation, \(y’ = 2x.\)
This differential equation has at least two solutions: \(y = {x^2}\) and \(y = {x^2} – 1.\)
Observe that the difference between the two possible solutions is the constant, \( – 1.\)
Let \(k,\) be the constant in the solution. Then, the solution can be written as \(y = {x^2} + k.\)
For all values of \(k,\) the differential equation remains the same as \(y = 2x.\)
The graphs of the solution for different values of \(k\) ranging from \(-2\) to \(2\) are shown below. As shown in the graph, any solution of the form \(y = {x^2} + k,\) is a solution as well for the given derivative. This holds for all values of k because the derivative of \({x^2} + k\) is \(2x.\) Hence, any solution of the form \(y = {x^2} + k\) can be the solution of \(y’ = 2x.\) Such a solution is called the general solution of the differential equation.

Particular Solution of an Ordinary Differential Equation (ODE)

The particular solution of a differential equation is achieved by substituting the extra values provided in the form of initial or boundary conditions. This solution without any arbitrary constants is called the particular solution.

Methods to Find Solutions

There are two methods to find the solution for a differential equation. They are:
1. Integrating factor
2. Separation of variables

Solving using Integrating Factor

An integrating factor is a function that is used to solve differential equations. This method is employed when the first order differential equation is of the form:
\(\frac{{dy}}{{dx}} + Py = Q\)
Where \(P\) and \(Q\) are functions of the independent variable \(x.\)
Let \(I\) be the integrating factor which is also a function of \(x.\)
Multiplying the differential equation on both sides, we get,
\(I\left( {\frac{{dy}}{{dx}} + Py} \right) = IQ\)
\(I\frac{{dy}}{{dx}} + IPy = IQ\)
Integrating on both sides,
\(\int {\left( {I\frac{{dy}}{{dx}} + IPy} \right)} dx = \int {IQ\,dx} \)
Using product rule, we get,
\(Iy = \int {IQ\,dx} \)
As \(I\) and \(Q\) are functions of \(x,\) the general solution to the differential equation \(\frac{{dy}}{{dx}} + Py = Q\) is found by integrating \(IQ\) with respect to the independent variable \(x.\) Then, the equation is rearranged to solve for \(y.\)

Solving using Separation of Variables

As the name suggests, separation of variables can be used in differential equations where the independent variable \(x\) and \(dx\) and the dependent variable \(y\) and \(dy\) can be rearranged on either side of the equation.
Follow the following steps to solve a differential equation using separation of variables:
Step 1: Move all the \(x\) and \(dx\) to one side of the equation, and move all the \(y\) and \(dy\) to the other side of the equation.
Step 2: Integrate one side with respect to \(x\) and integrate the other side with respect to \(y.\) Include the arbitrary constant, say \(k.\)
Step 3: Simplify and solve for \(y.\)

Linear Differential Equations

Linear differential equations are the commonly used type of differential equations. A linear equation is a differential equation of the form:
\({a_n}\left( t \right){y^n}\left( t \right) + {a_{n – 1}}\left( t \right){y^{n – 1}}\left( t \right) + {a_{n – 2}}\left( t \right){y^{n – 2}}\left( t \right) + … + {a_1}\left( t \right){y^1}\left( t \right) + {a_0}\left( t \right){y^0}\left( t \right) = f\left( t \right)\)
where,
\({a_0}\left( t \right),\,{a_1}\left( t \right)…{a_n}\left( t \right)\) and \(f\left( t \right) \to \) coefficients
The coefficients can be any function such as zero, nonzero, constant, non-constant, linear or nonlinear.
Only the function \(y\left( t \right)\) and its derivatives are used to ascertain if a function is linear.
The first order linear differential equation is written as\(\frac{{dy}}{{dx}} + Ay = B,\) where, \(P\) and \(Q\) are constants or functions of the independent variable \(y\).
Any differential equation that cannot be written in the above format is called a nonlinear differential equation.

Solved Examples on the Basics of Differential Equation

Q.1. Find the order and degree of the differential equation: \({\left( {\frac{{dy}}{{dx}}} \right)^5} – 2x = 3\sin \,x – \sin \,y\)
Ans: Given: \({\left( {\frac{{dy}}{{dx}}} \right)^5} – 2x = 3\sin \,x – \sin \,y\)
Here,

The highest derivative is \(1.\) Hence, the order of the differential equation \(= 1\)

The exponent of the highest derivative is \(5\) Hence, the degree of the differential equation \( = 5\)

Q.2. Solve the ordinary differential equation (ODE) for \(x\left( t \right).\)
\(\frac{{dx}}{{dt}} = 5x – 3\)
Ans: Given: \(\frac{{dx}}{{dt}} = 5x – 3\)
\(\frac{{dx}}{{5x – 3}} = dt\)
Integrating on both sides,
\(\int {\frac{{dx}}{{\left( {5x – 3} \right)}}} = \int {dt} \)
\(\frac{1}{5}\log \left| {5x – 3} \right| = t + k\)
\(5x – 3 = \pm \exp \left( {5t + 5k} \right)\)
\(x = \pm \frac{1}{5}{e^{5t + 5k}} + \frac{3}{5}\)
\(x = \pm \frac{1}{5}{e^{5t}}{e^{5k}} + \frac{3}{5}\)
Let, \( \pm \frac{1}{5}{e^{5k}} = C.\) Then,
\(x\left( t \right) = C{e^{5t}} + \frac{3}{5}\)

Q.3. Find the general solution of \(y + \cos \left( {x + y} \right) + \left( {x – y + \cos \left( {x + y} \right)} \right)\frac{{dy}}{{dx}} = 0.\)
Ans: Given: \(y + \cos \left( {x + y} \right) + \left( {x – y + \cos \left( {x + y} \right)} \right)\frac{{dy}}{{dx}} = 0.\)
Integrating on both sides,
\(\int {y + \cos \left( {x + y} \right) + \left( {x – y + \cos \left( {x + y} \right)} \right)\frac{{dy}}{{dx}}dx} = \int 0 \)
\(xy + \sin \left( {x + y} \right) – \frac{1}{2}{y^2} = k\)

4. Show that \(y = c\,\sin \,2x + 3\,\cos \,2x\) is the general solution of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0.\)
Ans: We know that: \(y = c\,\sin \,2x + 3\,\cos \,2x\)
First derivative: \(\frac{{dy}}{{dx}} = 2c\,\cos \,2x – 6\,\sin \,2x\)
Second derivative: \(\frac{{{d^2}y}}{{d{x^2}}} = \, – 4c\,\sin \,2x – 12\,\cos \,2x\)
Given: \(\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\)
Substituting the known values,
\({\rm{L}}{\rm{.H}}{\rm{.S}} = \, – 4c\,\sin \,2x – 12\,\cos \,2x + 4\left( {c\,\sin \,2x + 3\,\cos \,2x} \right)\)
\( = \, – 4c\,\sin \,2x – 12\,\cos \,2x + 4c\,\sin \,2x + 12\,\cos \,2x\)
\(=\,0\)
\( = {\rm{R}}{\rm{.H}}{\rm{.S}}\)
Hence proved, \(y = c\,\sin \,2x + 3\,\cos \,2x\) is the general solution of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0.\)

Q.5. Find the particular solution of the differential equation \(y’ = 2x\) passing through the point \(\left( {2,\,7} \right).\)
Ans: Given: \(y’ = 2x\)
Integrating on both sides,
\(\int {\frac{{dy}}{{dx}}dx} = \int {2x\,dx} \)
\(y = {x^2} + k\)
Substitute \(\left( {3,\,5} \right)\) in \(y.\)
Here,
\(x = 3\)
\(y = 5\)
\(5 = {3^2} + k\)
\(k = 5 – 9\)
\(k = \, – 4\)
\(\therefore \) The particular solution of the differential equation \(y’ = 2x\) passing through the point \(\left( {2,\,7} \right)\) is \(y = {x^2} – 4.\)

Summary

The article defines ordinary differential equations (ODE) are equations that involve a function in one variable and its derivative. The term ordinary is only used to differentiate it from partial differential equations. The article also explains identifying the order and degree of differential equations and then defines their types. It defines linear differential equations. There are two types of solutions for differential equations – general solution and particular solution. While the general solution has an arbitrary constant, the particular solution is achieved using the additional values provided in the problem. The article also provides solved examples to apply the studied concepts and calculations.

Frequently Asked Questions (FAQs) on Basics Differential Equation

Q.1. What is a differential equation in simple terms?
Ans: A differential equation is an equation that has a function in one variable and its derivative. For example, \(\frac{{{d^2}y}}{{d{t^2}}} = 4\,{y^3}{x^2}\) is a differential equation, where, \(\frac{{{d^2}y}}{{d{t^2}}}\) is the second order derivative. The solution of a differential equation provides the dependent variable \(y\) in terms of the independent variable \(x.\)

Q.2. What are differential equations used for?
Ans: Differential equations are used in various fields such as economics, physics, chemistry, and biology too! They are used to calculate exponential growth and decay, increase in population over time, and the compounded interest of one’s savings.

Q.3. What is the formula of the differential equation?
Ans: The formula of an ordinary differential equation is \(\frac{{dy}}{{dx}} = f\left( x \right).\) Note that this is also a first-order differential equation.

Q.4. What is the general solution of a differential equation?
Ans: The general solution of a differential equation is the relation between the independent variable \(x,\) and the dependent variable \(y,\) which is received after solving to remove the differential terms in that equation. This relation is called the general solution of differential equations, and it also includes an arbitrary constant.

Q.5. What is a first-order differential equation?
Ans: A first-order differential equation is a polynomial that has the order of \(1.\) This means that the first derivative is the highest in that equation. It takes the form \(f\left( {t,\,x,\,x’} \right) = 0.\) Observe that the order of the equation is \(1,\) and is represented by \(x’.\) Example: \(x\frac{{dy}}{{dx}} + y\,\cos \,x = {x^2}\sin \,x\)

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