Methods of Solving First Order First Degree Differential Equation

Methods of solving first order, first degree differential equation: Differential equations have several real life applications such as in computing the movement or flow of electricity, analysing the to and fro motion of an object such as a pendulum, and visualising the progression of diseases in a graphical form in medical field. A first order and first-degree differential equation can be written as \(f(x,y)dx + g(x,y)dy = 0\) or \(\frac{{dy}}{{dx}} = \frac{{f(x,y)}}{{g(x,y)}}\) or \(\frac{{dy}}{{dx}} = \phi (x,y)\), where \(f(x,y)\) and \(g(x,y)\) are the functions of \(x\) and\(y\).

Differential equations of these kinds are not always solvable. A first order and first-degree differential equation can be solved only if it fits into certain category of standard forms. In this post, we’ll go through some of these common standard forms and methods of solving First Order, First Degree Differential Equation.

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Methods of Solving First Order First Degree Differential Equation

Before going into the details on how to solve First Order, First Degree Differential Equation, let us first look at the type of differential equations

Types of Differential Equations

We shall discuss several techniques of obtaining solutions for the following types of differential equations such as:

• Differential equations of the form \(\frac{d y}{d x}=f(x)\)
• Differential equations of the form \(\frac{d y}{d x}=f(y)\)
• Differential equations in variable separable form
• Differential equations reducible to variable separable form
• Homogeneous differential equations
• Linear differential equations

Differential Equations of the Type \(\frac{d y}{d x}=\boldsymbol{f}(\boldsymbol{x})\)

We have

\(\frac{d y}{d x}=f(x)\)

\(\Rightarrow d y=f(x) d x\)

Integrating both sides, we obtain,

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

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

This gives the general solution of the differential equation. 

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Differential Equations of the Type \(\frac{d y}{d x}=f(y)\)

Consider

\(\frac{d y}{d x}=f(y)\)

\(\Rightarrow \frac{d x}{d y}=\frac{1}{f(y)}\) , provided that \(f(y) \neq 0\)

\(\Rightarrow d x=\frac{1}{f(y)} d y\)

Integrating both sides, we obtain, 

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

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

Equations in Variable Separable Form

The variables are said to be separable if the differential equation can be written as \(f(x) d x=g(y) d y\) integrating both sides can be used to solve such equations. Thus, its solution is given by

\(\int f(x) d x=\int g(y) d y+C\), where \(C\) is an arbitrary constant.

Equations Reducible to Variable Separable Form

By substituting \(a x+b y+c=v\) for \(\frac{d y}{d x}=f(a x+b y+c)\) differential equations of the type \(\frac{d y}{d x}=f(a x+b y+c)\) can be reduced to variable separable form.

As \(a x+b y+c=v\)

\(\Rightarrow a+b \frac{d y}{d x}=\frac{d v}{d x}\)

\(\Rightarrow \frac{d y}{d x}=\frac{1}{b}\left[\frac{d v}{d x}-a\right]\)

Since \(\frac{d y}{d x}=f(a x+b y+c)\)

\(\Rightarrow \frac{1}{b}\left[\frac{d v}{d x}-a\right]=f(v)\)

\(\Rightarrow \frac{d v}{d x}=b f(v)+a\)

\(\Rightarrow \int \frac{d v}{b f(v)+a}=\int d x\)

On integrating the above expression, we get the required solution.

Homogeneous Differential Equation

A homogeneous differential equation is one in which the first-order and first-degree differential equation is expressed in the form,

\(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\)

Where \(f(x, y)\) and \(g(x, y)\) are homogeneous functions of the same degree.

By substituting \( y=v x\) such equations can be simplified to variable separable form.

The given differential equation can be written as,

\(\frac{d y}{d x}=\frac{x^{n} f(y / x)}{x^{n} g(y / x)}=\frac{f(y / x)}{g(y / x)}=F\left(\frac{y}{x}\right)\) (say)

Let \(\frac{y}{x}=v \Rightarrow y=v x\)

\(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

Substituting these values in  \(\frac{d y}{d x}=F\left(\frac{y}{x}\right)\), we get,

\(v+x \frac{d v}{d x}=F(v)\)

\(\Rightarrow \frac{1}{F(v)-v} d v=\frac{d x}{x}\)

On integration, we get \(\int \frac{1}{F(v)-v} d v=\int \frac{d x}{x}+C\) where \(C\) is an arbitrary constant.

After integration,\(v\) is replaced by \(\frac{y}{x}\) to get the complete solution. 

What are Exact Differential Equations?

A differential equation of type,

\(M(x, y) d x+N(x, y) d y=0\)

 is called an exact differential equation if there exists a function of two variables \(u(x, y)\) with continuous partial derivatives such that

\(d u(x, y)=M(x, y) d x+N(x, y) d y\)

The general solution of an exact equation is given by,

\(u(x, y)=C\)

Where, \(C\) is an arbitrary constant.

Condition for Exactness

Let functions \(M(x, y)\) and \(N(x, y)\) have continuous partial derivatives in a certain domain \(D.\) The differential equation \(M(x, y) d x+N(x, y) d y=0\) is an exact equation if and only if 

\(\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\)

Linear Differential Equations

If the dependent variable \((y)\) and its derivative occur only in first degree, the differential equation is linear. The general form of a first-order linear differential equation is

\(\frac{d y}{d x}+P y=Q\)…..(i)

Where, \(P\) and \(Q\) are functions of\(x\) or constants

Example: 

\(\frac{d y}{d x}+x y=x^{3}\)

\(x \frac{d y}{d x}+2 y=x^{3}\)

\(\frac{d y}{d x}+2 y=\sin x\)

This sort of differential equation is solved by multiplying it by an integrating factor. This is because the LHS of \((i)\) becomes the exact differential of a function when multiplied by this factor.

Multiplying both sides of \((i)\) by \(e^{\int P d x}\), we get,

\(e^{\int P d x}\left(\frac{d y}{d x}+P y\right)=Q e^{\int P d x}\)

\(\Rightarrow \frac{d}{d x}\left(y e^{\int p d x}\right)=Q e^{\int p d x}\)

On integrating both sides with respect to \(x\), we get,

\(y e^{\int P d x}=\int Q e^{\int P d x} d x+C\)…….(ii)

This is the required solution, where \(C\) is the constant of integration. 

The solution (ii) may also be written as,

\(y(\mathrm{I} . \mathrm{F} .)=\int Q(\mathrm{I} . \mathrm{F} .) d x+C\)

How to Solve?

These steps may be used to solve a linear differential equation.

Step 1 :  Write the differential equation in the form \(\frac{d y}{d x}+P y=Q\)

Step 2 : Obtain \(P\) and \(Q\)

Step 3 : Find integrating factor (I.F.) given by \(I . F .=e^{\int P d x}\)

Step 4 : Multiply both sides of equation in Step \(1\) by I.F.

Step 5 : Integrate both sides of the equation obtained in step \(4\) with respect to \(x\) to obtain the solution,

\(y(\mathrm{I} . \mathrm{F} .)=\int Q(\mathrm{I} . \mathrm{F} .) d x+C\)

Linear Differential Equations of the Form \(\frac{d x}{d y}+\boldsymbol{R} \boldsymbol{x}=\boldsymbol{S}\)

Given,

\(\frac{d x}{d y}+R x=S\)

Here,

\(y \rightarrow\) independent variable

\(x \rightarrow\) dependent variable

\(R, S \rightarrow\) functions of \(y\) or constants.

How to Solve?

The following steps are used to solve these types of equations.

Step 1 :  Write the differential equation in the form \(\frac{d x}{d y}+R x=S\)

Step 2 : Obtain \(R\) and \(S\)

Step 3 : Find (I.F.) by using \(I.F.=e^{\int R d y}\)

Step 4 : Multiply both sides of the differential equation in Step \(1\) by I.F.

Step 5 : Integrate both sides of the equation obtained in step \(4\) with respect to \(y\) to obtain the solution given by

\(x\) (I.F.) \(=\int S\) (I.F.) \(d y+C\)

Where,\(c\) is the constant of integration.

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Solved Examples – Methods of Solving First Order, First Degree Differential Equation

Q.1. Solve the following differential equation. \((x+2) \frac{d y}{d x}=x^{2}+4 x-9, x \neq-2\)
Ans: Given \((x+2) \frac{d y}{d x}=x^{2}+4 x-9, x \neq-2\)
\(\Rightarrow \frac{d y}{d x}=\frac{x^{2}+4 x-9}{x+2}\)
Integrating both sides, we get,
\(\int d y=\int \frac{x^{2}+4 x-9}{x+2} d x\)
\(\Rightarrow \int d y=\int\left(x+2-\frac{13}{x+2}\right) d x\)
\(\Rightarrow y=\frac{x^{2}}{2}+2 x-13 \log |x+2|+C\)
Hence, \(y=\frac{x^{2}}{2}+2 x-13 \log |x+2|+C, x \in R-{2}\) is the solution of the given differential equation.

Q.2. Solve: \(\frac{d y}{d x}+y=1\)
Ans: Given: \(\frac{d y}{d x}+y=1\)
\(\Rightarrow \frac{d y}{d x}=1-y\)
\(\Rightarrow \frac{d x}{d y}=\frac{1}{1-y}\)
\(\Rightarrow d x=\frac{1}{1-y} d y\)
Integrating both sides, we get, 
\(\int d x=\int \frac{1}{1-y}-d y\)
\(\Rightarrow x=-\log |1-y|+C\) is the required solution. 

Q.3. Solve the following differential equation \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
Ans: Given: \(\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y\)
\(\Rightarrow \frac{d y}{d x}=\sin (x+y)\) ……(i)
Let \(x+y=v\) ……(ii)
\(\Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x}\)
\(\Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1\) ……..(iii)
Substituting the values from equation (ii) and (iii) in equation (i) we have,
\(\frac{d v}{d x}-1=\sin v\)
\(\Rightarrow \frac{d v}{d x}=1+\sin v\)
\(\Rightarrow \frac{1}{1+\sin v} d v=d x\)
\(\Rightarrow \int \frac{1}{1+\sin v} d v=\int d x\)
\(\Rightarrow \int d x=\int \frac{1-\sin v}{1-\sin ^{2} v} d v\)
\(\Rightarrow \int d x=\int \frac{1-\sin v}{\cos ^{2} v} d v\)
\(\Rightarrow \int d x=\int\left(\sec ^{2} v-\tan v \sec v\right) d v\)
\(\therefore x = \tan (x + y) – \sec (x + y) + C\) is the required solution.

Q.4. Solve the differential equation \((x+y) d y+(x-y) d x=0\) given that \(y=1\) when \(x=1\)
Ans: Given \((x+y) d y+(x-y) d x=0\)
\(\Rightarrow \frac{d y}{d x}=-\frac{x-y}{x+y}\)
\(\Rightarrow \frac{d y}{d x}=\frac{y-x}{x+y}\) …… (i)
Since each of the functions \(y-x\) and \(x+y\) is a homogeneous function of degree \(1\). equation (i) is a homogeneous differential equation.
Let \(y=vx\)
\(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in equation (i), we have,
\(v+x \frac{d v}{d x}=\frac{v x-x}{x+v x}\)
\(\Rightarrow v+x \frac{d v}{d x}=\frac{v-1}{v+1}\)
\(\Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v\)
\(\Rightarrow x \frac{d v}{d x}=\frac{v-1-v^{2}-v}{v+1}\)
\(\Rightarrow x \frac{d v}{d x}=-\left(\frac{v^{2}+1}{v+1}\right)\)
\(\Rightarrow \frac{v+1}{v^{2}+1} d v=-\frac{d x}{x}\) [ By seperating the variables]
\(\Rightarrow \int \frac{v+1}{v^{2}+1} d v=-\int \frac{d x}{x}\)
\(\Rightarrow \int \frac{v}{v^{2}+1} d v+\int \frac{1}{v^{2}+1} d v=-\int \frac{d x}{x}\)
\(\Rightarrow \int \frac{v}{v^{2}+1} d v+\int \frac{1}{v^{2}+1} d v=-\int \frac{d x}{x}\)
\(\Rightarrow \frac{1}{2} \int \frac{2 v}{v^{2}+1} d v+\int \frac{1}{v^{2}+1^{2}} d v=-\int \frac{d x}{x}\)
\(\Rightarrow \frac{1}{2} \log \left(v^{2}+1\right)+\tan ^{-1} v=-\log |x|+C\)
\(\Rightarrow \log \left(v^{2}+1\right)+2 \log |x|+2 \tan ^{-1} v=2 C\)
\(\Rightarrow \log \left(v^{2}+1\right)+\log x^{2}+2 \tan ^{-1} v=k\) Where \(k=2 C\)
\(\Rightarrow \log \left\{ {\left( {{v^2} + 1} \right){x^2}} \right\} + 2{\tan ^{ – 1}}v = k\)
\( \Rightarrow \log \left\{ {\left( {\left( {\frac{{{y^2}}}{{{x^2}}}} \right) + 1} \right) + {x^2}} \right\} + 2{\tan ^{ – 1}}\left( {\frac{y}{x}} \right) = k\,\left[ {\because v = yx} \right]\)
\(\Rightarrow \log \left(x^{2}+y^{2}\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)=k\)…..(ii)
It is given that \(y=1\) when \(x=1\) Putting \(x=1, y=1\) in (ii), we get,
\(\log 2+2 \tan ^{-1}(1)=k\)
\(\Rightarrow k=\log 2+2\left(\frac{\pi}{4}\right)\)
\(\therefore k=\frac{\pi}{2}+\log 2\)
Substituting the value of \(k\) in (ii) we get,
\(\log \left(x^{2}+y^{2}\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)=\frac{\pi}{2}+\log 2\)
Hence, \(\log \left(x^{2}+y^{2}\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)=\frac{\pi}{2}+\log 2, x \neq 0\) is the required solution.

Q.5. Solve the differential equation: \(\frac{d y}{d x}+\frac{y}{2 x}=3 x^{2}\)
Ans: Given \(\frac{d y}{d x}+\frac{y}{2 x}=3 x^{2}\)…..(i)
This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\)
Where,
\(P=\frac{1}{2 x}\)
\(Q=3 x^{2}\)
\(\therefore\) I. F. \(=e^{\int P d x}\)
\(=e^{\int \frac{1}{2 x} d x}\)
\(=e^{\frac{1}{2} \log x}\)
\(=e^{\log x^{\frac{1}{2}}}\)
I.F. \(=x^{\frac{1}{2}}\)
Multiplying both sides of (i) by I.F \(=\sqrt{x}\) we get,
\(\sqrt{x} \frac{d y}{d x}+\frac{1}{2 \sqrt{x}} y=3 x^{\frac{5}{2}}\)
Integrating both sides with respect to \(x\) we get,
\(y \sqrt{x}=\int 3 x^{\frac{5}{2}} d x+C\) [ Using \(y\) (I.F.) \(=\int Q\) (I.F.) \(d x+C\)
\(\Rightarrow y \sqrt{x}=3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+C\)
\(\Rightarrow y \sqrt{x}=\frac{6}{7} x^{\frac{7}{2}}+C\)
\(\therefore y=\frac{6}{7} x^{3}+C x^{-\frac{1}{2}}\) is the required solution.

Q.6. Solve: \(y d x-\left(x+2 y^{2}\right) d y=0\)
Ans: Given \(y d x-\left(x+2 y^{2}\right) d y=0\)
\(\Rightarrow \frac{d y}{d x}=\frac{y}{x+2 y^{2}}\)
\(\Rightarrow \frac{d x}{d y}=\frac{x+2 y^{2}}{y}\)
\(\Rightarrow \frac{d x}{d y}+\left(-\frac{1}{y}\right) x=2 y\) ……..(i)
This is a linear differential equation of the form \(\frac{d x}{d y}+R x=S\)
Where,
\(R=-\frac{1}{y}\)
\(S=2 y\)
\(\therefore\) I.F. \(=e^{\int R d y}\)
\(=e^{\int \frac{-1}{y} d y}\)
\(=e^{-\log y}\)
\(=e^{\log y^{-1}}\)
I.F. \(=y^{-1}\)
Multiplying both sides of (i) by I.F \(=y^{-1}\), we obtain,
\(\frac{1}{y} \frac{d x}{d y}-\frac{1}{y^{2}} x=2\)
Integrating both sides with respect to \(y\) we get,
\(x \times \frac{1}{y} = \int {2dy + C} \) [Using: \(x\left( {I.F.} \right) = \int {S\left( {I.F.} \right)dy + C} \)]

Summary

A first order and first-degree differential equation is of the form: \(f(x, y) d x+g(x, y) d y=0\) or \(\frac{d y}{d x}=\frac{f(x, y)}{g(x, y)}\) or \(\frac{{dy}}{{dx}} = \phi \left( {x,\,y} \right).\) These equations can be solved by identifying their standard form. If the differential equation can be separated as \(f(x) d x=g(y) d y\), then we can solve it by variable separable method. If it is homogeneous then it can be reduced to variable separable form to find its solution. Linear differential equation of the form, \(\frac{d y}{d x}+P y=Q\) or \(\frac{d x}{d y}+R x=S,\) can be solved using the integrating factor given by \(I . F .=e^{\int p d x}\) or \({\rm{I}}{\rm{. F}}{\rm{. = }}{e^{\int R dy}}\) Various solved examples and FAQs are provided to make you understand how to use these formulas.

Frequently Asked Questions (FAQs)

Q.1. How many ways can you solve first order differential equations?
Ans: First order differential equation can be solved using variable separable method. Homogeneous differential equations can be reduced to variable separable form and then be solved. If the differential equation is of the form \(\frac{d y}{d x}+P y=Q\) and \(\frac{d x}{d y}+R x=S\) then we can solve it using an integrating factor \(I . F .=e^{\int P d x}\)

Q.2. Which is the first order and first-degree differential equation?
Ans: A first order and first degree differential equation has
1. The highest order derivative as \(\frac{d y}{d x}\)
2. There is no product of derivatives and dependent variable
3. the power of the derivative and dependent variable is \(1\)
It can be written as
\(M(x, y) d x+N(x, y) d y=0\)
or \(\frac{d y}{d x}=\frac{M(x, y)}{N(x, y)}\)
or \(\frac{d y}{d x}=\phi(x, y)\)
where \(M(x, y)\) and \(N(x, y)\) are the functions of \(x\) and \(y\)

Q.3. How do you find the order of a differential equation?
Ans: The order of a differential equation is the order of the highest order derivative in the equation.
Example: In the equation \(\frac{d^{3} y}{d x^{3}}+3 \frac{d y}{d x}+2 y=e^{x}\) the order of highest order derivative is \(3\). So, it is a differential equation of order \(3\).

Q.4. How do you solve first order homogeneous differential equations?
Ans: To find the general solution of homogeneous differential equation, follow the steps given below.
Step 1: Substitute \(y=v x\)
Step 2: Reduce the given equation to variable separable form.
Step 3: Find the solution of this equation.

Q.5. What is a first order linear equation?
Ans: If the dependent variable \((y)\) and its derivative occur only in first degree, then the differential equation is said to be first order. All such first order differential equations are also linear. General form of a first-order linear differential equation is given by,
\(\frac{d y}{d x}+P y=Q\)

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