Methods of Solving Differential Equation: A differential equation is an equation that contains one or more functions with its derivatives. It is primarily used in physics, engineering, biology, etc. The differential equation’s primary purpose is to study solutions that satisfy the equations. The solution of the differential equation is the relationship between the variables included, which satisfies the given differential equation.

There are two types of solutions of differential equations, namely, the general solution and the particular solution. The general and the particular solutions of differential equations make use of some steps of integration to solve the equations. In this article, let us learn more about methods to solve the differential equation such as variable separable method, homogenous differential equation, and linear differential equation with the help of indefinite integration formula.

Solution of Differential Equations

The solution of a differential equation is an expression for the dependent variable in terms of the independent variable which satisfies the given differential equation, so the solution can be classified into two types;

• General solution
• Particular solution

General Solution

A solution that consists of as many arbitrary constants is called a general solution.

Particular Solution

The solution is obtained by assigning particular values to the arbitrary constants in the general solution of the differential equation, and then the resulting solution is a particular solution. The result of eliminating one arbitrary constant yields to a first-order differential equation, and eliminating two arbitrary constants yields to a second-order differential equation.

Methods to Solve First Order, First Degree Differential Equations

There are several methods to solve the first order, first-degree differential equation;

• Solution by inspection method
• Variable separable method
• Homogenous equations
• First-order, linear equations

Inspection Method

If the differential equation is of the form \(f\left( {{f_1}\left( {x,y} \right)} \right)d\left( {{f_1}\left( {x,y} \right)} \right) + \varphi \left( {{f_2}\left( {x,y} \right)} \right)d\left( {{f_2}\left( {x,y} \right)} \right) + \cdots \ldots = 0,\) then each term can be integrated separately. So, the solution of the differential equation can be obtained by using the inspection method. There are a few results that will help us to find the solution of differential equation

Type	Solution
\(d\left( {x + y} \right)\)	\(dx + dy\)
\(d\left( {xy} \right)\)	\(xdy + ydx\)
\(d\left( {\frac{x}{y}} \right)\)	\(\frac{{ydx – xdy}}{{{y^2}}}\)
\(d\left( {\frac{y}{x}} \right)\)	\(\frac{{xdy – ydx}}{{{x^2}}}\)
\(d\left( {\frac{{{x^2}}}{y}} \right)\)	\(\frac{{2xydx – {x^2}dy}}{{{y^2}}}\)

Variable Separable Method

The first order, first-degree differential equation, is of the form
\(\frac{{dy}}{{dx}} = f\left( {x,y} \right)….\left( i \right)\)
If \(f\left( {x,y} \right)\) is expressed as the product of \(f\left( x \right)g\left( y \right),\) where \(f\left( x \right)\) is a function of \(x\) and \(g\left( y \right)\) is a function of \(y;\) then the differential equation is said to be variable separable type. Then, we can write \(\left( i \right)\) as
\(\frac{{dy}}{{dx}} = f\left( x \right)g\left( y \right)\)
We can follow three simple steps to find the solution using the variable separable method.
Step 1: Move all \(y\) terms including \(dy\) to one side of the equation and all the \(x\) terms including \(dx\) to the other side.
Step 2: Integrate one side with respect to \(y\) and the other side with respect to \(x,\) and add the constant of integration.
Step 3: Simplify the final result to the simplest form.

Homogenous Differential Equations

A homogeneous differential equation is an equation that contains a derivative and a function with a set of variables. The function \(f\left( {x,y} \right)\) in a homogeneous differential equation is a homogeneous function such that \(f\left( {tx,ty} \right) = {t^n}f\left( {x,y} \right),\) for non-zero constant \(t.\) So, the general form of the homogenous differential equation is of the form \(f\left( {x,y} \right)dy + g\left( {x,y} \right)dx = 0.\)
Note: The homogenous differential equation does not contain any constant term within the equation.
Some of the examples of a homogenous differential equation are:
\( \bullet \frac{{dy}}{{dx}} = \frac{{\left( {x + y} \right)}}{{\left( {x – y} \right)}}\)
\( \bullet \frac{{dy}}{{dx}} = \frac{{x\left( {x – y} \right)}}{{{y^2}}}\)

Steps to Solve Homogenous Differential Equations

To solve the homogenous differential equation, we follow these steps.
Case 1:
If the differential equation of the type \(\frac{{dy}}{{dx}} = F\left( {x,y} \right) = g\left( {\frac{y}{x}} \right).\)
Step 1: Substitute \(y = vx\) in the given differential equation.
Step 2: Differentiate with respect to \(x,\) then \(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}.\) Now substitute the values of \(y\) and \(\frac{{dy}}{{dx}}\) in the given differential equation, then we have \(v + x\frac{{dv}}{{dx}} = g\left( v \right)\)
\( \Rightarrow x\frac{{dv}}{{dx}} = g\left( v \right) – v\)
Step 3: Separating the variables, then \(\frac{{dv}}{{g\left( v \right) – v}} = \frac{{dx}}{x}.\)
Step 4: Apply integration on both sides, then \(\int {\frac{{dv}}{{g\left( v \right) – v}}} dv = \int {\frac{{dx}}{x}} + C.\)
Step 5: After the integration, substitute the value of \(v\) from Step \(1.\)
Case 2:
If the differential equation of the type \(\frac{{dx}}{{dy}} = F\left( {x,y} \right) = g\left( {\frac{x}{y}} \right).\) These kinds of homogenous differential equations can be solved by putting \(x = vy,\) after that, follow the same five steps which we applied for case \(1.\)

Linear Differential Equation

The linear differential equation involves a variable, a derivative of this variable, and a few other functions. The standard equation of a linear differential equation is \(\frac{{dy}}{{dx}} + Py = Q,\) where \(P\) and \(Q\) are functions of \(x.\) Similarly, \(\frac{{dx}}{{dy}} + {P_1}x = {Q_1}\) is also the standard equation of a linear differential equation. Here \({P_1}\) and \({Q_1}\) are functions of \(y\)

General Solution of Linear Differential Equation

Type 1: The general solution of differential equation of the type \(\frac{{dy}}{{dx}} + Py = Q,\) where \(P\) and \(Q\) are functions of \(x\) is \(y\left( {I.F} \right) = \int {\left( {Q\left( {I.F} \right)dx} \right) + C,} \) where Integrating Factor \(\left( {I.F} \right) = {e^{\int {P\,dx} }}.\)
Type 2: The general solution of differential equation of the type \(\frac{{dx}}{{dy}} + Px = Q,\) where \(P\) and \(Q\) are functions of \(y\) is \(x\left( {I.F} \right) = \int {\left( {Q\left( {I.F} \right)dy} \right) + C,} \) where Integrating Factor \(\left( {I.F} \right) = {e^{\int {P\,dy} }}.\)

Steps to Solve Linear Differential Equation

The general solution of the linear differential equation can be solved by using the following three simple steps;
Step 1: Simplify the given differential equation in the form of \(\frac{{dy}}{{dx}} + Py = Q\) or \(\frac{{dx}}{{dy}} + Px = Q,\) where \(P\) and \(Q\) are functions of \(x\) or \(P\) and \(Q\) are functions of \(y\)
Step 2: Find the integrating factor by using \(\left( {I.F} \right) = {e^{\int {P\,dx} }}\) or \(\left( {I.F} \right) = {e^{\int {P\,dy} }}.\)
Step 3: Finally obtain the general solution by using the formula:
\(y\left( {I.F} \right) = \int {\left( {Q\left( {I.F} \right)dx} \right) + C} \) or is \(x\left( {I.F} \right) = \int {\left( {Q\left( {I.F} \right)dy} \right) + C} \)

Solved Examples on Methods of Solving Differential Equations

Q.1. Find the general solution of the following differential equation
\(xdy – \left( {y + 2{x^2}} \right)dx = 0\)
Ans: Given: \(xdy – \left( {y + 2{x^2}} \right)dx = 0\)
\( \Rightarrow \frac{{dy}}{{dx}} – \frac{y}{x} = 2x\)
On comparing with the linear differential equation \(\frac{{dy}}{{dx}} + Py = Q,\) we have
\(P = – \frac{1}{x},Q = 2x\)
Now, the integrating factor is \(\left( {I.F} \right) = {e^{\int { – \frac{1}{x}} dx}}\)
\( = {e^{ – \log x}}\)
\(\therefore I.F = \frac{1}{x}\)
Therefore, the general solution is \(y \times \frac{1}{x} = \int {\left( {2x \times \frac{1}{x}} \right)} dx + c\)
\( \Rightarrow \frac{y}{x} = \int 2 dx + c\)
\( \Rightarrow \frac{y}{x} = 2x + c\)
\(\therefore y = 2{x^2} + xc\) is the required general solution of the given differential equation.

Q.2. Find the integrating factor following the differential equation
\(\frac{{dy}}{{dx}} + \sec x \cdot y = \tan x\)
Ans: Given: \(\frac{{dy}}{{dx}} + \sec x \cdot y = \tan x\)
On comparing with the linear differential equation \(\frac{{dy}}{{dx}} + Py = Q,\) We have
\(P = \sec x,Q = \tan x\)
Now, the integrating factor is \(\left( {I.F} \right) = {e^{\int {\sec xdx} }}\)
\( = {e^{\ln \left| {\sec x + \tan x} \right|}}\)
\(\therefore I.F = \left| {\sec x + \tan x} \right|\)

Q.3. Show that the following differential equation \(\left( {x + y} \right)\frac{{dy}}{{dx}} = \left( {x + 2y} \right)\) is a homogeneous differential equation.
Ans: Given: \(\left( {x + y} \right)\frac{{dy}}{{dx}} = \left( {x + 2y} \right)\)
To prove that the above differential equation is a homogeneous differential equation, substitute \(x = tx,\) and \(y = tx\)
So, We have \(F\left( {x,y} \right) = \frac{{\left( {x + 2y} \right)}}{{\left( {x + y} \right)}}\)
\( \Rightarrow F\left( {tx,ty} \right) = \frac{{\left( {tx + 2ty} \right)}}{{\left( {tx + ty} \right)}}\)
\( \Rightarrow F\left( {tx,ty} \right) = \frac{{t\left( {x + 2y} \right)}}{{t\left( {x – y} \right)}} = {t^0}F\left( {x,y} \right)\)
\(\therefore F\left( {tx,ty} \right) = {t^0}F\left( {x,y} \right)\)
Hence, the given differential equation is homogeneous.

Q.4. Find the general solution of the following differential equation \(\frac{{dy}}{{dx}} = 6{y^2}x.\)
Ans: Given: \(\frac{{dy}}{{dx}} = 6{y^2}x.\)
It is clear that the given differential equation is variable separable as we can separate the variables.
Now, separating the variables, we get
\(\frac{1}{{{y^2}}}dy = 6xdx\)
\( \Rightarrow {y^{ – 2}}dy = 6xdx\)
Apply integration on both sides, then we get: \(\int {{y^{ – 2}}} dy = \int 6 xdx\)
\( \Rightarrow – \frac{1}{y} = \frac{{6{x^2}}}{2} + C\)
\(\therefore – \frac{1}{y} = 3{x^2} + C\) is the required general solution.

Q.5. Find the general solution of following differential equation \(\frac{{dy}}{{dx}} = \frac{{{x^2} + {y^2}}}{{xy}}.\)
Ans: Given: \(\frac{{dy}}{{dx}} = \frac{{{x^2} + {y^2}}}{{xy}}.\)
It is clear that the given the differential equation is a homogenous differential equation as it is of the form \(F\left( {\frac{y}{x}} \right)\)
Now, substitute \(y = vx\)
Differentiate with respect to \(x,\) then we get
\(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\)
Therefore, \(v + x\frac{{dv}}{{dx}} = {v^{ – 1}} + v\)
\( \Rightarrow x\frac{{dv}}{{dx}} = {v^{ – 1}}\)
Now, separate the variables, then \(vdv = \frac{1}{x}dx\)
Apply integration on both sides; then we get \(\int v dv = \int {\frac{1}{x}} dx\)
\( \Rightarrow \frac{{{v^2}}}{2} = \ln \left( x \right) + C\)
\( \Rightarrow \frac{{{v^2}}}{2} = \ln \left( x \right) + \ln \left( k \right)\) where \(C = \ln \left( k \right)\)
\( \Rightarrow \frac{{{v^2}}}{2} = \ln \left( {kx} \right)\)
\( \Rightarrow v = \pm \sqrt {\left( {2\ln \left( {kx} \right)} \right)} \)
\( \Rightarrow \frac{y}{x} = \pm \sqrt {\left( {2\ln \left( {kx} \right)} \right)} \) [resubstitute \(v = \frac{y}{x}\)]
\(\therefore y = \pm x\sqrt {\left( {2\ln \left( {kx} \right)} \right)} ,\) which is a required general solution.

Summary of Methods of Solving Differential Equations

The solution of the differential equation is the relationship between the variables included, which satisfies the given differential equation. There are two types of solutions for differential equations such as the general solution and the particular solution. These solutions of differential equations make use of some steps of integration to solve the equations. Later, explained the types of differential equations followed by methods of the differential equation such as variable separable method, homogenous method, and linear differential equation along with some solved examples are explained.

Frequently Asked Questions (FAQs)

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