General and Particular Solutions of a Differential Equation

General and Particular Solutions of a Differential Equation: The functions of a differential equation often describe physical values, whereas the derivatives express the rate of change of the physical quantities. A differential equation is a connection that exists between a function and its derivatives. These equations appear in a wide range of applications, including Physics, Chemistry, Biology, Anthropology, Geology, and Economics. As a result, a thorough understanding of differential equations has become critical in all current scientific inquiries.

In this article, we will learn about the general solution and the method of finding the particular solution from the general solution of the differential equation. We will also learn to solve the initial value problems and a few ways to find the solution of the first order and first-degree differential equations.

Solution of a Differential Equation

A differential equation solution is a connection that satisfies the differential equation between the variables involved. When such a relationship and the derivatives produced from it are replaced in a differential equation, the left and right sides are equal.

For example, \(y = {e^x}\) is a solution of the differential equation \(\frac{{dy}}{{dx}} = y\).

Consider the differential equation

\(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\, \ldots \ldots .(i)\)

Also, consider the relation \(y = A\,\cos \,x + B\,\sin \,x\, \ldots \ldots (ii)\)

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

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

\(\frac{{dy}}{{dx}} = \,- A\,\sin \,x + B\,\cos \,x\)

Again differentiating this with respect to \(x\), we get

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

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} =\, – y\)

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

This demonstrates that \(y = A\,\cos \,x + B\,\sin \,x\) satisfies the differential equation \((i)\) indicating that it is a solution of the differential equation provided in \((i)\).

 It is simple to verify that \(y = 3\,\cos \,x + 2\,\sin \,x,\,y = A\,\cos \,x,\,y = B\,\sin \,x\), and other are also solutions of the differential equation given in \((i)\). We discovered that the solution \(y = 3\,\cos \,x + 2\,\sin \,x\) has no random constants, whereas the solutions \(y = A\,\cos \,x,\,y = B\,\sin \,x\) have just one. The solution \(y = A\,\cos \,x + B\,\sin \,x\) contains two arbitrary constants, so it is known as the general solution of \((i)\) whereas all other solutions are particular solutions.

General Solution

The general solution of the differential equation is one that comprises as many arbitrary constants as the order of the differential equation. Thus, a general solution of \(ntℎ\) order differential equation has \(n\) arbitrary constants. For example, \(y = A\,\cos \,x + B\,\sin \,x\) is the general solution of the second order differential equation \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\). 

The general solution depicts an \(n\)−parameter family of curves geometrically. A one-parameter family of curves is defined by the general solution of the differential equation \(\frac{{dy}}{{dx}} = 3{x^2}\), which is \(y = {x^3} + c\) where \(c\) is an arbitrary constant, as shown in the image below.

Particular Solution of a Differential Equation

• A particular solution of a differential equation is a solution achieved by giving specified values to the arbitrary constants in the general solution.
• The criteria for computing the values of arbitrary constants might be given to us as an initialvalue problem or as boundary conditions.

Singular Solution

The singular solution of a differential equation is also a particular solution, but it cannot be produced from the general solution by defining the values of the arbitrary constants.

Initial Value Problems

We have seen that a first order differential equation represents a one-parameter family of curves, a second order differential equation represents a two-parameter family of curves, and so on. Therefore, if we wish to specify a particular member of such a family of curves, then in addition to the differential equation we require some other conditions for the specification of the parameter(s). 

These conditions are generally prescribed by assigning values to the unknown function (dependent variable) and its various order derivatives at some point of the domain of definition of independent variable.

For example, one-parameter family of curves given by \(y = 2{x^2} + C\) is represented by the differential equation:

\(\frac{{dy}}{{dx}} = 4x\)

In order to specify a particular member, say \(y = 2{x^2} + 3\) of this family, we require the differential equation: 

\(\frac{{dy}}{{dx}} = 4x\) and the condition \(y(1) = 5\)

i.e. \(y = 5\) when \(x = 1\)

Putting \(x = 1\), and \(y = 5\) in the general solution of the differential equation \(\frac{{dy}}{{dx}} = 4x\), we get \({\rm{C = 5 – 2 = 3}}\).

Similarly, two-parameter family of curves given by \(y = 3{x^2} + ax + b\) is represented by the differential equation:

\(\frac{{{d^2}y}}{{d{x^2}}} – 6 = 0\)

Now, if we want to specify a particular member, say \(y = 3{x^2} – 2x + 1\) of this family, then, we require the differential equation:

\(\frac{{{d^2}y}}{{d{x^2}}} – 6 = 0\) and two conditions, namely,

\(y{\rm{(0) = 1}}\)

\({y^\prime }(0) =\, – 2\)

It follows from the above discussion that to specify a particular member of a family of curves, we require 

• the differential equation representing the given family of curves 
• the values of dependent variable
•  its various order derivatives at some point of the domain of definition. 

These values are generally prescribed at only one point of the domain of definition of independent variable and are generally referred to as initial values or initial conditions.

Solution of First Order and First Degree Differential Equations

 A first order and first degree differential equation can be written as 

\(f\left( {x,\,y} \right)dx + g\left( {x,\,y} \right)dy = 0\)

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

where \(f(x,\,y)\) and \(g(x,\,y)\) are the functions of \(x\) and \(y\).

Solving these kinds of equations isn’t always doable. Only when this sort of differential equation fits into the category of some standard forms it is possible to solve it. Let’s discuss some of the standard forms and method of obtaining their solutions.

Differential Equation of the type \(\frac{{dy}}{{dx}} = f\left( x \right)\)

Consider

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

\( \Leftrightarrow dy = f(x)dx\)

 Integrating both sides, we obtain,

\(\int d y = \int f (x)dx + C\)

\( \Rightarrow y = \int f (x)dx + C\)

This is the general solution of the differential equation.

Differential Equation of the type \(\frac{{dy}}{{dx}} = f(y)\)

Consider

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

\( \Rightarrow \frac{{dx}}{{dy}} = \frac{1}{{f(y)}}\), provided \(f(y) \ne 0\)

\( \Rightarrow dx = \frac{1}{{f(y)}}dy\)

 Integrating both sides, we get,

\(\int d x = \int {\frac{1}{{f(y)}}} dy + C\)

\( \Rightarrow x = \int {\frac{1}{{f(y)}}} dy + C\), which gives the general solution of the differential equation.

Solved Examples – General and Particular Solutions of a Differential Equation

Q.1. Show that \(y = Ax + \frac{B}{x},\,x \ne 0\) is a solution of the differential equation.
Ans: We have
\(y = Ax + \frac{B}{x},\,x \ne 0\)
Differentiating both sides with respect to \(x\), we get
\(\frac{{dy}}{{dx}} = A – \frac{B}{{{x^2}}}\)
Differentiating with respect to \(x\), we get
\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{{2B}}{{{x^3}}}\)
Substituting the values of \(y,\,\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) in \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} – y\), we get
\({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} – y = {x^2}\left( {\frac{{2B}}{{{x^3}}}} \right) + x\left( {A – \frac{B}{{{x^2}}}} \right) – \left( {Ax + \frac{B}{x}} \right) = \frac{{2B}}{x} + Ax – \frac{B}{x} – Ax – \frac{B}{x} = 0\)
Thus, the function \(y = Ax + \frac{B}{x}\) satisfies the differential equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} – y = 0\).
Hence, \(y = Ax + \frac{B}{x}\) is a solution of the given differential equation.

Q.2. Show that the function \(y = (A + Bx){e^{3x}}\) is a solution of the equation \(\frac{{{d^2}y}}{{d{x^2}}} – 6\frac{{dy}}{{dx}} + {\rm{9}}y = {\rm{0}}\).
Ans: Wehave
\(y = (A + Bx){e^{3x}}\, \ldots \ldots (i)\)
Differentiating \((i)\) with respect to \(x\), we get
\(\frac{{dy}}{{dx}} = B{e^{3x}} + 3{e^{3x}}(A + Bx)\, \ldots \ldots (ii)\)
Differentiating \((ii)\) with respect to \(x\), we get
\(\frac{{{d^2}y}}{{d{x^2}}} = 6B{e^{3x}} + 9{e^{3x}}(A + Bx)\)
\(\therefore \,\frac{{{d^2}y}}{{d{x^2}}} – 6\frac{{dy}}{{dx}} + 9y\)
\( = \left\{ {6B{e^{3x}} + 9{e^{3x}}(A + Bx)} \right\} – 6\left\{ {B{e^{3x}} + 3{e^{3x}}(A + Bx)} \right\} + \left\{ {9(A + Bx){e^{3x}}} \right\}\)
\( = 6B{e^{3x}} + 9A{e^{3x}} + 9Bx{e^{3x}} – 6B{e^{3x}} – 18A{e^{3x}} – 18Bx{e^{3x}} + 9A{e^{3x}} + 9Bx{e^{3x}}\)
\(\therefore \,\frac{{{d^2}y}}{{d{x^2}}} – 6\frac{{dy}}{{dx}} + 9y = 0\)
Thus, \(y = (A + Bx){e^{3x}}\) satisfies the given differential equation.
Hence, it is a solution of the given differential equation.

Q.3. Verify that the function \(y = {C_1}{e^{ax}}\,\cos \,bx + {C_2}{e^{ax}}\,\sin \,bx\), where \({C_1},\,{C_2}\) are arbitrary constants is a solution of the differential equation \(\frac{{{d^2}y}}{{d{x^2}}} – 2a\frac{{dy}}{{dx}} + \left( {{a^2} + {b^2}} \right)y = 0\).
Ans: We have
\(y = {C_1}{e^{ax}}\,\cos \,bx + {C_2}{e^{ax}}\,\sin \,bx\, \ldots \ldots (i)\)
Differentiating both sides with respect to \(x\), we get
\(\frac{{dy}}{{dx}} = {C_1}\left\{ {a{e^{ax}}\,\cos \,bx – b{e^{ax}}\,\sin \,bx} \right\} + {C_2}\left\{ {a{e^{ax}}\,\sin \,bx + b{e^{ax}}\,\cos \,bx} \right\}\)
\( \Rightarrow \frac{{dy}}{{dx}} = ay + b\left\{ { – {C_1}{e^{ax}}\,\sin \,bx + {C_2}{e^{ax}}\,\cos \,bx} \right\}\, \ldots ..(ii)\)
Differentiating with respect to \(x\), we get
\(\frac{{{d^2}y}}{{d{x^2}}} = a\frac{{dy}}{{dx}} + b\left\{ { – a{C_1}{e^{ax}}\,\sin \,bx – b{C_1}{e^{ax}}\,\cos \,bx + a{C_2}{e^{ax}}\,\cos \,bx – b{C_2}{e^{ax}}\,\sin \,bx} \right\}\)
\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = a\frac{{dy}}{{dx}} + ab\left\{ { – {C_1}{e^{ax}}\,\sin \,bx + {C_2}{e^{ax}}\,\cos \,bx} \right\} – {b^2}\left\{ {{C_1}{e^{ax}}\,\cos \,bx + {C_2}{e^{ax}}\,\sin \,bx} \right\}\)
\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = a\frac{{dy}}{{dx}} + a\left\{ {\frac{{dy}}{{dx}} – ay} \right\} – {b^2}y\) [Using \((i)\) and \((ii)\)]
\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} – 2a\frac{{dy}}{{dx}} + \left( {{a^2} + {b^2}} \right)y = 0\)
Hence, the given function is a solution of the given differential equation.

Q.4. Verify that the function defined by \(y = \sin \,x – \cos \,x,\,x \in R\) is a solution of the initial value problem \(\frac{{dy}}{{dx}} = \sin \,x + \cos \,x,\,y(0) =\, – 1\).
Ans: Given, \(y = \sin \,x – \cos \,x,\,x \in R\)
\( \Rightarrow \frac{{ay}}{{dx}} = \cos \,x + \sin \,x\) which is the given differential equation.
Thus, \(y = \sin \,x – \cos \,x\) satisfies the differential given equation and hence it is a solution. Also, when \(x = 0,\,y = \sin \,0 – \cos \,0 = 0 – 1 = – 1\) i.e. \(y(0) =\, – 1\).
Hence, \(y = \sin \,x – \cos \,x\) is a solution of the given initial value problem.

Q.5. Show that \(y = {x^2} + 2x + 1\) is the solution of the initial value problem \(\frac{{{d^3}y}}{{d{x^3}}} = 0,\,y(0) = 1,\,{y^\prime }(0) = 2,\,{y^{\prime \prime }}(0) = 2\).
Ans: Given, \(y = {x^2} + 2x + 1\)
\( \Rightarrow \frac{{dy}}{{dx}} = 2x + 2\)
\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 2\)
\( \Rightarrow \frac{{{d^3}y}}{{d{x^3}}} = 0\), which is the given differential equation. 
Thus, \(y = {x^2} + 2x + 1\) satisfies the given differential equation.
Hence, it is a solution of the given differential equation.
Also, at \(x = 0\), we have
\(y(0) = {0^2} + 2(0) + 1 = 1,\,{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = 2(0) + 2 = 2\)
And \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 0}} = 2\)
\( \Rightarrow y(0) = 1,\,{y^\prime }(0) = 2\) and \({y^{\prime \prime }}(0) = 2\)
Hence, \(y = {x^2} + 2x + 1\) is the solution of the initial value problem.

Q.6. Solve the following differential equation.
\(\frac{{dy}}{{dx}} = \frac{{3{e^{2x}} + 3{e^{4x}}}}{{{e^x} + {e^{ – x}}}}\)
Ans: Consider
\(\frac{{dy}}{{dx}} = \frac{{3{e^{2x}} + 3{e^{4x}}}}{{{e^x} + {e^{ – x}}}}\)
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{3{e^{2x}}\left( {1 + {e^{2x}}} \right)}}{{{e^x} + \frac{1}{{{e^x}}}}}\)
\( = \frac{{3{e^{2x}}\left( {1 + {e^{2x}}} \right)}}{{\frac{{{e^{2x}} + 1}}{{{e^x}}}}}\)
\( = \frac{{3{e^{3x}}\left( {1 + {e^{2x}}} \right)}}{{\left( {1 + {e^{2x}}} \right)}}\)
\( \Rightarrow \frac{{dy}}{{dx}} = 3{e^{3x}}\)
\( \Rightarrow dy = 3{e^{3x}}dx\)
On integrating both sides of the above equation
\( \Rightarrow \int d y = 3\int {{e^{3x}}} dx\)
\( \Rightarrow y = 3\left( {\frac{{{e^{3x}}}}{3}} \right) + C\)
\( \Rightarrow y = {e^{3x}} + C\), which is the required solution.

Q.7. Solve the initial value problem \(\frac{{dy}}{{dx}} + 2{y^2} = 0,\,y(1) = 1\) and find the corresponding solution curve. 
Ans: Given, \(\frac{{dy}}{{dx}} + 2{y^2} = 0\)
\( \Rightarrow \frac{{dy}}{{dx}} = \,- 2{y^2}\)
\( \Rightarrow \frac{{dx}}{{dy}} =\, – \frac{1}{{2{y^2}}}\)
Integrating both sides with respect to \(y\), we get
\(\int d x =\, – \int {\frac{1}{{2{y^2}}}} dy\)
\( \Rightarrow x = \frac{1}{{2y}} + C\, \ldots \ldots (i)\)
It is given that \(y(1) = 1\) i.e. \(y = 1\) when \(x = 1\).
Putting \(x = 1\) and \(y = 1\) in \((i)\), we get
\(1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}\)
Putting \(C = 12\) in \((i)\), we get
\(x = \frac{1}{{2y}} + \frac{1}{2}\)
\( \Rightarrow y = \frac{1}{{2x – 1}}\), which is the required solution curve.

Summary

Solutions of differential equations are mainly of two types, general solution and particular solution. The differential equation’s available solution comprises as many arbitrary constants as the order of the differential equation. A particular solution of a differential equation is achieved by assigning specific values to the arbitrary constants in the general solution. The method of solving these arbitrary constants involves the initial value problems, and we have learned the technique of finding the general solution of the first order and first-degree differential equations by using integration.

Frequently Asked Questions (FAQs)

Q.1. What is general and particular solution of differential equation?
Ans: The general solution of the differential equation is one that comprises as many arbitrary constants as the order of the differential equation.
The particular solution of a differential equation is a solution computed by giving specified values to the arbitrary constants in the general solution.

Q.2. How are Slope fields related to general and particular solutions to differential equations? 
Ans: A visual depiction of a differential equation of the form \(\frac{{dy}}{{dx}} = f(x,\,y)\) is called a slope field. There is a tiny line segment whose slope equals the value of \(f\) at each sample point \(\left( {x,\,y} \right)\).

Q.3. What is the difference between general and particular solution of a differential equation?
Ans: A particular solution is just one that solves the entire ODE; a general solution, on the other hand, is the total solution of an ODE, which is the sum of complementary and particular solutions.

Q.4. How many solutions does a differential equation have?
Ans: A differential equation, as we’ve seen, often has an unlimited number of solutions. A general solution is the name given to such a solution. There is just one solution to a related initial value problem. A particular solution is one in which there are no unknown constants left.

Q.5. What is the meaning of a solution of differential equation?
Ans: The function \(y = f\left( x \right)\) that satisfies the differential equation when substituted along with its derivatives is called the solution of the differential equation.

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