Formation of Differential Equation: The functions of a differential equation often describe physical values, whereas the derivatives express the rate of change of those physical quantities. A differential equation is a connection that exists between a function and its derivatives.

These appear in a wide range of applications, including physics, chemistry, biology, anthropology, geology, and economics. Hence, a thorough understanding of differential equations has become critical in all current scientific inquiries.  In this article, we will learn some basic concepts related to differential equations such as order and degree of differential equations, and their formation too.

What are Differential Equations?

A differential equation is any equation having an independent variable, a dependent variable, and differential coefficients of the dependent variable with respect to the independent variable.

Example: \(\frac{{dy}}{{dx}} = 2xy,\,\frac{{{d^2}y}}{{d{x^2}}} = 4x\) and \(\frac{{dy}}{{dx}} = \sin \,x + \cos \,x\)

Classification of Differential Equations

We can classify the differential equations on the basis of the number of independent variables present in them.

1. Ordinary Differential Equation:

The “Ordinary Differential Equation,” or ODE, is a mathematical equation with just one independent variable and one or more derivatives with respect to the variable. As a result, the ordinary differential equation is represented as a relation with one independent variable \(x\) and one real dependent variable \(y\), as well as some of its derivatives \({y^\prime },\,{y^{\prime \prime }},\, \ldots {y^n}\) with regard to \(x\).

Example: \(\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} = 3y\cos \,x\)

2. Partial Differential Equation:

“Partial Differential Equation”, or PDE, is a mathematical equation involving two or more independent variables, an unknown function (depending on the variables), and partial derivatives of the unknown function with respect to the independent variables.

Example: \(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 4xy\frac{{{\partial ^2}u}}{{\partial {y^2}}} + u = 2{\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}}} \right)^2} + 4\frac{{{\partial ^3}u}}{{\partial x\partial {y^2}}} = 10x\)

Order of a Differential Equation

The order of a differential equation is the order of the highest order derivative in the equation.

Example: In the equation \(\frac{{{d^3}y}}{{d{x^3}}} + 3\frac{{dy}}{{dx}} + 2y = {e^x}\), the order of highest order derivative is \(3\). So, it is a differential equation of order \(3\). 

The equation \(\frac{{{d^2}y}}{{d{x^2}}} – 6{\left( {\frac{{dy}}{{dx}}} \right)^3} – 4y = 0\) is of order \(2\) because the order of highest order derivative in it is \(2\).

Note: A differential equation’s order is always a positive integer.

Degree of a Differential Equation

The degree of a differential equation is the power of the highest order derivative in the equation. In other terms, when written as a polynomial with differential coefficients, the power of the highest order derivative is called the degree of the differential equation.

For example, in the differential equation, \(x{\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^2} + {\left( {\frac{{dy}}{{dx}}} \right)^4} + {y^2} = 0\), the highest order derivative is \(3\), and its power is \(2\). Hence, degree of this differential equation is \(2\).

Linear and Non-linear Differential Equations

 A differential equation is said to be linear, if it is expressible in the form:

\({A_0}\frac{{{d^n}y}}{{d{x^n}}} + {A_1}\frac{{{d^{n – 1}}y}}{{d{x^{n – 1}}}} + {A_2}\frac{{{d^{n – 2}}y}}{{d{x^{n – 2}}}} + \ldots + {A_{n – 1}}\frac{{dy}}{{dx}} + {A_n}y = Q\)

Where, \({A_0},\,{A_1},\,{A_2},\, \ldots ,\,{A_{n – 1}},\,{A_n}\) and \(Q\) are either constants or functions of the independent variable \(x\).

To identify a given equation as a linear differential equation, it should satisfy the following conditions:

• The derivatives and dependent variable are raised to the first power.
• There is no product of derivatives and dependent variable in the equation.
• The coefficients of the various components are either constants or functions of the independent variable. 

It is a non-linear differential equation otherwise. 

It follows from the above definition that a differential equation will be non-linear differential equation if:

• Its degree is more than \(1\).
• any differential coefficient has more than one exponent.
• the exponent of the dependent variable is more than one.
• there are products that contain the dependent variable and its differential coefficients.

Example:

• \({\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^3} – 6{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} – 4y = 0\) is a non-lineardifferential equation, because its degree is \(3\), more than one. 
• \(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\) is a non-lineardifferential equation, because the differential coefficient \(\frac{{dy}}{{dx}}\) has exponent \(2\).  
• \(\left( {{x^2} + {y^2}} \right)dx – 2xy\;dy = 0\) is a non-linear differential equation, because the exponent of dependent variable y is \(2\) and it involves the  product of \(y\) and \(\frac{{dy}}{{dx}}\). 
• \(\frac{{{d^2}y}}{{d{x^2}}} – 5\frac{{dy}}{{dx}} + 6y = \sin \,x\) is a linear differential equation of order \(2\) and degree \(1\).

Formation of Differential Equations

Consider the family of curves given by, \(y = A{e^x}\), where A is the parameter. For different values of \(A\), we obtain different equations of the same family.

Differentiating \(y = A{e^x}\) with respect to \(x\), we get,

\(\frac{{dy}}{{dx}} = A{e^x}\)

On eliminating the parameter \(A\) between \(y = A{e^x}\) and \(\frac{{dy}}{{dx}} = A{e^x}\), we get,

\(\frac{{dy}}{{dx}} = y\)

This is the differential equation of the family of curves represented by the equation \(y = A{e^x}\). 

Thus, by eliminating one arbitrary constant, a differential equation of first order is obtained. In other words, we can say that a one-parameter family of curves is represented by a first-order differential equation.

Now, consider a two-parameter family of curves given by,

\(y = A\cos \,2x + B\sin \,2x\,……(i)\)

where, \(A\) and \(B\) are  are arbitrary constants.

Differentiating \((i)\) with respect to \(x\), we get

\(\frac{{dy}}{{dx}} = – 2A\sin \,2x + 2B\cos \,2x\,……(ii)\)

Differentiating \((ii)\) with respect to \(x\), we get

\(\frac{{{d^2}y}}{{d{x^2}}} = – 4A\cos \,2x – 4B\sin \,2x\,……(iii)\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = – 4(A\cos \,2x + B\sin \,2x)\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = – 4y\) [Using \((i)\)]

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\)

Here, note that by eliminating two arbitrary constants, a differential equation of second order is obtained. In other words, a two-parameter family of curves is represented by a second-order differential equation.

Similarly, by eliminating three arbitrary constants, a differential equation of third order is obtained. We can say that, a three-parameter family of curves is represented by a third-order differential equation. 

Hence, it can be deduced that if an equation contains n arbitrary constants, a differential equation of \({n^{th}}\) order may be generated by removing the n arbitrary constants. This means that an \(n\)-parameter family of curves is represented by a differential equation of \({n^{th}}\) order.

In order to formulate a differential equation, follow the steps given below:

• Step 1: Write the given equation which involves a dependent variable \(y\), an independent variable \(x\), and arbitrary constants.
• Step 2: Find the number of arbitrary constants in the given equation. Let it be \(n\).
• Step 3: Differentiate the equation in step \(1, n\) times with respect to \(x\).
• Step 4: Solve the \(n + 1\) equations (equations obtained in step \(3\), and that in step \(1\)) in such a way that there are zero arbitrary constants left.

The resulting equation is the required differential equation.

Solved Examples – Formation of Differential Equation

Q.1. Form the differential equation for the family of curves represented \(c{(y + c)^2} = {x^3}\), where \(c\) is a parameter.
Ans:

Given: The equation of the family of curves is:

\(c{(y + c)^2} = {x^3}\,……(i)\)

Clearly, it is a one-parameter family of curves, so we shall get a differential equation of first order.

Differentiating \((i)\) with respect to \(x\), we get:

\(2c(y + c)\frac{{dy}}{{dx}} = 3{x^2}\,……(ii)\)

Dividing \((i)\) by \((ii)\), we get,

\(\frac{{c{{(y + c)}^2}}}{{2c(y + c)\left( {\frac{{dy}}{{dx}}} \right)}} = \frac{{{x^3}}}{{3{x^2}}}\)

\( \Rightarrow y + c = \frac{{2x}}{3}\frac{{dy}}{{dx}}\)

\(\therefore \,c = \frac{{2x}}{3}\frac{{dy}}{{dx}} – y\)

Substituting this value of \(c\) in \((i)\), we get,

\(\left( {\frac{{2x}}{3}\frac{{dy}}{{dx}} – y} \right){\left( {\frac{2}{3}x\frac{{dy}}{{dx}}} \right)^2} = {x^3}\)

\( \Rightarrow \frac{4}{9}{\left( {\frac{{dy}}{{dx}}} \right)^2}\left( {\frac{{2x}}{3}\frac{{dy}}{{dx}} – y} \right) = x\)

\( \Rightarrow \frac{8}{{27}}x{\left( {\frac{{dy}}{{dx}}} \right)^3} – \frac{4}{9}{\left( {\frac{{dy}}{{dx}}} \right)^2}y = x\)

\( \Rightarrow 8x{\left( {\frac{{dy}}{{dx}}} \right)^3} – 12y{\left( {\frac{{dy}}{{dx}}} \right)^2} = 27x\) is the required differential equation.

Q.2. Form the differential equation corresponding to \({y^2} = a(b – x)(b + x)\) by eliminating parameters \(a\) and \(b\).
Ans:

Given: The equation of the family of curves is

\({y^2} = a(b – x)(b + x)\)

\( \Rightarrow {y^2} = a\left( {{b^2} – {x^2}} \right)\, \ldots \ldots (i)\)

There are two arbitrary constants. So, we differentiate it two times to get a differential equation of second order.

Differentiating \((i)\) with respect to \(x\), we get 

\(2y\frac{{dy}}{{dx}} = – 2ax\)

\( \Rightarrow y\frac{{dy}}{{dx}} = – ax\,……..(ii)\)

Differentiating \((ii)\) with respect to \(x\), we get,

\(y\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = – a\)

\(\therefore \,a =  – \left\{ {y\frac{{{d^2}y}}{{d{x^2}}} + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\} \; …….(iii)\)

Substituting the value of a obtained from \((iii)\) in \((ii)\), we get,

\(x\left\{ {y\frac{{{d^2}y}}{{d{x^2}}} + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\} = y\frac{{dy}}{{dx}}\)

Q.3. Find the differential equation of all the circles in the first quadrant which touch the  coordinate axes.
Ans:

The equation of the family of circles in the first quadrant which touch the  coordinate axes is \({(x – a)^2} + {(y – a)^2} = {a^2}\, \ldots \ldots (i)\)

Here, \(a\) is the only parameter. So, differentiate once.

Differentiating \((i)\) with respect to \(x\), we get,

\(2(x – a) + 2(y – a)\frac{{dy}}{{dx}} = 0\)

\( \Rightarrow x – a + (y – a)\frac{{dy}}{{dx}} = 0\)

\( \Rightarrow a = \frac{{x + y\frac{{dy}}{{dx}}}}{{1 + \frac{{dy}}{{dx}}}}\)

\(\therefore \,a = \frac{{x + py}}{{1 + p}}\), where \(p = \frac{{dy}}{{dx}}\)

Substituting the value of \(a\) in \((i)\), we get:

\({\left( {x – \frac{{x + py}}{{1 + p}}} \right)^2} + {\left( {y – \frac{{x + py}}{{1 + p}}} \right)^2} = {\left( {\frac{{x + py}}{{1 + p}}} \right)^2}\)

\( \Rightarrow {(xp – py)^2} + {(y – x)^2} = {(x + py)^2}\)

\( \Rightarrow {(x – y)^2}{p^2} + {(x – y)^2} = {(x + py)^2}\)

\( \Rightarrow {(x – y)^2}\left( {{p^2} + 1} \right) = {(x + py)^2}\)

\(\therefore \,{(x – y)^2}\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\} = {\left( {x + y\frac{{dy}}{{dx}}} \right)^2}\) is the required differential equation.

Q.4. Form the differential equation of the family of ellipses having foci on the y-axis and centre at the origin.
Ans: The equation of the family of ellipses having a centre at the origin and foci on \(y\)-axis is

\(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\), where \(b > a\,……..(i)\)

This is two parameter family of ellipses. So, we differentiate twice.

Differentiating with respect to \(x\), we get,

\(\frac{{2x}}{{{a^2}}} + \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\)

\( \Rightarrow \frac{x}{{{a^2}}} + \frac{y}{{{b^2}}}\frac{{dy}}{{dx}} = 0\,……..(ii)\)

Differentiating \((ii)\) with respect to \(x\), we get,

\(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{y}{{{b^2}}}\frac{{{d^2}y}}{{d{x^2}}} = 0\,……..(iii)\)

Multiplying throughout by \(x\), we get,

\(\frac{x}{{{a^2}}} + \frac{x}{{{b^2}}}{\left( {\frac{{dy}}{{dx}}} \right)^2} + \frac{{xx}}{{{b^2}}}\frac{{{d^2}y}}{{d{x^2}}} = 0\,……..(iv)\)

Subtracting \((ii)\) from \((iv)\), we get

\(\frac{1}{{{b^2}}}\left\{ {x{{\left( {\frac{{dy}}{{dx}}} \right)}^2} + xy\frac{{{d^2}y}}{{d{x^2}}} – y\left( {\frac{{dy}}{{dx}}} \right)} \right\} = 0\)

\(\therefore \,x{\left( {\frac{{dy}}{{dx}}} \right)^2} + xy\frac{{{d^2}y}}{{d{x^2}}} – y\frac{{dy}}{{dx}} = 0\) is the required differential equation.

Q.5. Obtain the differential equation of all circles of radius \(r\).
Ans:

The equation of the family of circles of radius \(r\) is \({(x – a)^2} + {(y – b)^2} = {r^2}\,……..(i)\)

where \(a\) and \(b\) are parameters.

Clearly equation \((i)\) contains two arbitrary constants. So, let us differentiate it two times with  respect to \(x\).

Differentiating \((i)\) with respect to \(x\), we get

\(2(x – a) + 2(y – b)\frac{{dy}}{{dx}} = 0\)

\( \Rightarrow (x – a) + (y – b)\frac{{dy}}{{dx}} = 0\,……..(ii)\)

Differentiating \((ii)\) with respect to \(x\), we get

\(1 + (y – b)\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0\,……..(iii)\)

\( \Rightarrow y – b = – \frac{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\,……..(iv)\)

Putting this value of \((y − b)\) in \((ii)\), we obtain

\(x – a = \frac{{\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}\frac{{dy}}{{dx}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}} \; ……..(v)\)

Substituting the values of \((x − a)\) from \((v)\) and \((y − b)\) from \((iv)\) in \((i)\), we get

\(\frac{{{{\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}}^2}{{\left( {\frac{{dy}}{{dx}}} \right)}^2}}}{{{{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}}} + \frac{{{{\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}}^2}}}{{{{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}}} = {r^2}\)

\(\therefore \,{\left\{ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\}^3} = {r^2}{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}\) is the required differential equation.

Summary

This article explains the basic concepts related to the differential equations such as their order and degree, linear and non-linear differential equations, and formation of a differential equation. They are formed by eliminating all the arbitrary constants given in their solution. To eliminate the arbitrary constants, the solution is differentiated as many times as the number of arbitrary constants present in the equation corresponding to the family of curves. The solved examples in this article demonstrate how to form the differential equation from the given solution.

Frequently Asked Questions (FAQs)

Q.1. What is the process of formation of differential equations? 
Ans: If the given solution contains \(n\) arbitrary constants, then it has to be differentiated n times to obtain the \(n\) equations. Then, using these equations, and the given equation eliminate all the arbitrary constants. The equation so obtained is the required differential equation.

Q.2. Why is it called a differential equation?
Ans: We call it as a differential equation because they are equations involving a function and its derivatives. These derivatives are obtained by differentiating the function. Few examples of differential equations: \(\frac{{dy}}{{dx}} = 2xy,\,\frac{{{d^2}y}}{{d{x^2}}} = 4x,\,\frac{{dy}}{{dx}} = \sin \,x + \cos \,x\) etc.

Q.3. What is the order of a differential equation?
Ans: The order of a differential equation is the order of the equation’s highest order derivative.

For Example: In the equation \(\frac{{{d^3}y}}{{d{x^3}}} + 3\frac{{dy}}{{dx}} + 2y = {e^x}\), the order of highest order derivative is \(3\).

Q.4. What is non-linear differential equation?
Ans: A differential equation will be a non-linear differential equation if:

• It has more than one degree.
• Any differential coefficient has more than one exponent.
• The dependent variable’s exponent is more than one.
• There are products that contain the dependent variable and its differential coefficients.

Q.5. What is the general solution of a differential equation?
Ans: The solution of a differential equation is a relation between the variables involved which satisfies the differential equation. The general solution of the differential equation is one that comprises as many arbitrary constants as the order of the differential equation.

For example, \(y = A\cos \,x + B\sin \,x\) is the general solution of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\). We can check that this solution satisfies the differential equation.

Q.6. What is linear equation in differential equation?
Ans: A differential equation is said to be linear, if it is expressible in the form:

\({A_0}\frac{{{d^n}y}}{{d{x^n}}} + {A_1}\frac{{{d^{n – 1}}y}}{{d{x^{n – 1}}}} + {A_2}\frac{{{d^{n – 2}}y}}{{d{x^{n – 2}}}} + \ldots + {A_{n – 1}}\frac{{dy}}{{dx}} + {A_n}y = Q\)

Where, \({A_0},\,{A_1},\,{A_2},\, \ldots ,\,{A_{n – 1}},\,{A_n}\) and \(Q\) are either constants or functions of the independent variable \(x\).

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