Table of Contents

# Definition

An ordinary first order first degree differential equation is of the form

\[\frac{dy}{dx}=f(x,y)……….(1)\]

which can also be written as

\[Mdx+Ndy=0……….(2)\]

where M and N are functions of x and y or constants. All first order first degree differential equations can’t be solved. However, in the following series of articles we shall consider the following special types of equation of first order and first degree which can be solved by some standard method:

1. Equations in which variables are separable

2. Homogeneous equations

3. Exact equations

4. Linear equations

## Equations in which variables are separable

If the equation (1) or (2) can be put in the form

\[{{f}_{1}}(x)dx+{{f}_{2}}(y)dy=0\]

Then the equation can be solved easily by integrating each term separately. Thus the solution of the equation is

\[\int{{{f}_{1}}(x)dx+\int{{{f}_{2}}(y)dy=c}}\]

where c is an arbitrary constants.

Example 01 |

\[Solve:ydx+\left( 1+{{x}^{2}} \right){{\tan }^{-1}}xdy=0\]

**Solution:**

Given,

\[ydx+\left( 1+{{x}^{2}} \right){{\tan }^{-1}}xdy=0\]

\[\Rightarrow ydx=-\left( 1+{{x}^{2}} \right){{\tan }^{-1}}xdy\]

\[\Rightarrow \frac{dx}{\left( 1+{{x}^{2}} \right){{\tan }^{-1}}x}=-\frac{dy}{y}\]

\[\Rightarrow \frac{dx}{\left( 1+{{x}^{2}} \right){{\tan }^{-1}}x}+\frac{dy}{y}=0\]

Integrating, we get

\[\int{\frac{dx}{\left( 1+{{x}^{2}} \right){{\tan }^{-1}}x}+\int{\frac{dy}{y}}}=\log c\]

where c is an arbitrary constants.

\[\Rightarrow \int{\frac{d\left( {{\tan }^{-1}}x \right)}{\left( 1+{{x}^{2}} \right)}+}\int{\frac{dy}{y}}=\log c\]

\[\Rightarrow \log \left( {{\tan }^{-1}}x \right)+\log y=\log c\]

\[\Rightarrow \log \left( y{{\tan }^{-1}}x \right)=\log c\]

\[\therefore y{{\tan }^{-1}}x=c\]

is the required solution.

Example 02 |

\[Solve:\frac{dy}{dx}+1={{e}^{x+y}}\]

**Solution:**

Given

\[\frac{dy}{dx}+1={{e}^{x+y}}……….(1)\]

\[Let,x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}\]

Therefore equation (1) becomes

\[\frac{dz}{dx}={{e}^{z}}\Rightarrow {{e}^{-z}}dz=dx\]

Integrating, we get

\[\int{{{e}^{-z}}dz=\int{dx+c}}\]

\[\Rightarrow -{{e}^{-z}}=x+c\Rightarrow -{{e}^{-\left( x+y \right)}}=x+c\]

\[\therefore \frac{1}{{{e}^{\left( x+y \right)}}}+x+c=0\]

is the required solution.

## Homogeneous differential equations

A function f(x, y) is said to be homogeneous function of degree n in x and y if it can be expressed in the form

\[{{x}^{n}}\phi \left( \frac{y}{x} \right),or,{{y}^{n}}\mu \left( \frac{x}{y} \right)\]

When M and N are both homogeneous of same degree in x, y, then the equation M dx + N dy = 0 is called homogeneous equation. Hence a homogeneous equation can be put in the form

\[\frac{dy}{dx}=F\left( \frac{y}{x} \right)\]

Then putting y = vx where v is a function of x, the above equation becomes

\[v+x\frac{dv}{dx}=F(v)\]

\[\Rightarrow \frac{dv}{F(v)-v}=\frac{dx}{x}\]

Then integrating both sides, we get the solution in terms of v and x. finally replacing v by (y/x), we get the required solution.

Example |

\[Solve:\left( {{y}^{2}}-2xy \right)dx=\left( {{x}^{2}}-2xy \right)dy\]

**Solution:**

Here (y^{2} – 2xy) and (x^{2} – 2xy) both are homogeneous equation of same degree 2. The given equation can be written as

\[\frac{dy}{dx}=\frac{\left( {{y}^{2}}-2xy \right)}{\left( {{x}^{2}}-2xy \right)}……….(1)\]

\[Putting,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\]

Now the equation (1) becomes

\[v+x\frac{dv}{dx}=\frac{{{v}^{2}}{{x}^{2}}-2v{{x}^{2}}}{{{x}^{2}}-2v{{x}^{2}}}=\frac{{{v}^{2}}-2v}{1-2v}\]

\[\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-2v}{1-2v}-v\]

\[\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-2v-v+2{{v}^{2}}}{1-2v}=\frac{3v\left( v-1 \right)}{1-2v}\]

\[\Rightarrow 3\frac{dx}{x}=\frac{1-2v}{v\left( v-1 \right)}dv\]

\[\Rightarrow 3\frac{dx}{x}=-\frac{v+\left( v-1 \right)}{v\left( v-1 \right)}dv\]

\[\Rightarrow 3\frac{dx}{x}=-\left\{ \frac{1}{v-1}+\frac{1}{v} \right\}dv\]

Integrating we get

\[3\int{\frac{dx}{x}}=-\int{\left\{ \frac{1}{v-1}+\frac{1}{v} \right\}dv+\log c}\]

\[\Rightarrow 3\log x=-\log v-\log \left( v-1 \right)+\log c\]

\[\Rightarrow \log \left\{ {{x}^{3}}v\left( v-1 \right) \right\}=\log c\]

\[\Rightarrow {{x}^{3}}.\frac{y}{x}\left( \frac{y}{x}-1 \right)=c\]

\[\therefore xy\left( y-x \right)=c\]

is the required solution.

## Exact Differential equation

A first order first degree differential equation of the form

\[Mdx+Ndy=0……….(1)\]

Where both M and N are functions of x, y, is said to be exact, if there exist a function u(x, y) such that

\[Mdx+Ndy=du……….(2)\]

Then equation (1) becomes du = 0, which on integration gives u(x, y) = c, c being a constant.

Thus, u(x, y) = c is a solution of (1)

**For example,**

\[\log xdy+\frac{y}{x}dx=0,is-exact\]

\[\sin ce,\log xdy+\frac{y}{x}dx=d\left( y\log x \right)\]

Hence ylogx = c, c being a constant, is a general solution of the equation.

### Theorem:

The necessary and sufficient condition for the differential equation M dx + N dy = 0 to be exact is

\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]

### Method for solving M dx + N dy = 0 when it is exact

Step 1: First integrate the terms in M with respect to x, treating y as constant

Step 2: Then integrate, those terms of N which do not contain x, with respect to y.

Step 3: Lastly equate the sum of the results of Step 1 and Step 2 to a constant give the required solution.

Thus the solution of Mdx + Ndy = 0 is

ʃMdx (treating y as constant) + ʃ(terms of N not containing x)dy = c

Example |

\[Show,\left( 3x+4y+5 \right)dx+\left( 4x-3y+3 \right)dy=0\]

is an exact equation and hence solve it.

**Solution:**

\[Here,M=\left( 3x+4y+5 \right),N=\left( 4x-3y+3 \right)\]

\[\therefore \frac{\partial M}{\partial y}=4;\frac{\partial N}{\partial x}=4\]

\[\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]

So, the given equation is an exact equation. Hence the solution of the equation is

\[\int{\left( 3x+4y+5 \right)dx+\int{\left( -3y+3 \right)}}dy=c\]

\[\Rightarrow \frac{3{{x}^{2}}}{2}+4xy+5x-\frac{3{{y}^{2}}}{2}+3y=c\]

\[\therefore 3{{x}^{2}}-3{{y}^{2}}+8xy+10x+6y=2c=k\]

where k = 2c, is a constant

Thus the required solution.

## Linear differential equations

A first order first degree differential equation of the form

\[\frac{dy}{dx}+Py=Q……….(1)\]

Where P and Q are functions of x alone or constant, is known as first order linear equations.

To solve such type of equations, we use the integrating factor as

\[{{e}^{\int{Pdx}}}\]

Multiplying both sides of (1) by integrating factor we get

\[\frac{dy}{dx}.{{e}^{\int{Pdx}}}+P.y{{e}^{\int{Pdx}}}=Q{{e}^{\int{Pdx}}}\]

\[\Rightarrow \frac{d}{dx}\left( y{{e}^{\int{Pdx}}} \right)=Q{{e}^{\int{Pdx}}}\]

Integrating both sides we get

\[y{{e}^{\int{Pdx}}}=\int{Q{{e}^{\int{Pdx}}}dx+c}\]

is the required solution

Example |

\[Solve:{{\cos }^{2}}x\frac{dy}{dx}+y=\tan x\]

**Solution:**

The given equation can be written as

\[\frac{dy}{dx}+y{{\sec }^{2}}x=\tan x{{\sec }^{2}}x……….(1)\]

which is a linear equation in y

\[\therefore I.F.={{e}^{\int{{{\sec }^{2}}xdx}}}={{e}^{\tan x}}\]

Multiplying both sides of (1) by I.F. and integrating, we get

\[y{{e}^{\tan x}}=\int{\tan x{{\sec }^{2}}x}{{e}^{\tan x}}dx+c……….(2)\]

\[Let,\tan x=z\Rightarrow {{\sec }^{2}}xdx=dz\]

\[\therefore y{{e}^{\tan x}}=\int{z{{e}^{z}}dz+c}\]

\[\Rightarrow y{{e}^{\tan x}}=z\int{{{e}^{z}}dz-\int{\left\{ \frac{d}{dz}z.\int{{{e}^{z}}dz} \right\}}}dz\]+c

\[\Rightarrow y{{e}^{\tan x}}=z{{e}^{z}}-{{e}^{z}}+c\]

\[\Rightarrow y{{e}^{\tan x}}={{e}^{\tan x}}\left( \tan x-1 \right)+c\]

Is the required solution.

**In the upcoming articles we will learn about the methods thoroughly with lots of question solutions. Hope to see you then.**

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