# Differential Equation of First order and First degree

A differential equation of first order and first degree invokes x,y and $\displaystyle \frac { dy }{ dx }$ So it can be put in any one of the following forms :

$\displaystyle \frac { dy }{ dx } =f\left( x,y \right) \quad \\ or\quad f\left( x,y,\frac { dy }{ dx } \right) =0\\ or\quad f\left( x,y \right) dx+g\left( x,y \right) dy=0$

where f(x,y) and g(x,y) are obviously the function of x,y

Geometrical Interpretation of the differential equations of first order and first degree

The general form of a first order and first degree differential equation is

$\displaystyle f\left( x,y,\frac { dy }{ dx } \right) =0\qquad .......\left( i \right)$

We know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point given by  $\displaystyle \frac { dy }{ dx }$ so the equation in (i) cann be known as an equation which establishes the relationship between the coordinates of a point and the slope of the tangent i.e., $\displaystyle \frac { dy }{ dx }$ to the integral curve at that point.

Solving the differential equation given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field. All the curves represented by the general solution when taken together will give locus of the differential equation. Since there is one arbitrary constant in the general solution of the equation of first order, the locus of the equation can be said to be made up of single infinity of curves.

Solution of First order and first degree differential equation

As discussed earlier a first order and first degree differential equation can written as

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

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

It is not always possible to solve this type of equations. This solution of this type of differential equations. The solution of some standard forms and methods of obtaining their solutions.

Methods of solving a first order first degree differential equation

In this section we shall discuss several techniques of obtaining solutions of various types of differential equations.

Type I  Differential equations of the type $\displaystyle \frac { dy }{ dx } =f\left( x \right)$

To solve this type of differential equations we integrate both sides to obtain the general solution as discussed below:

$\displaystyle \frac { dy }{ dx } =f\left( x \right) \\ \Leftrightarrow dy=f\left( x \right)dx$

Integrating both sides, we obtain

$\displaystyle \int { dy } =\int { f\left( x \right) } dx+C\\ or\quad y=\int { f\left( x \right) } dx+C$

### Example

Solve

$\displaystyle \frac { dy }{ dx } =\frac { x }{ { x }^{ 2 }+1 }$

Solution:

$\displaystyle We\quad have\quad \frac { dy }{ dx } =\frac { x }{ { x }^{ 2 }+1 } \\ \Rightarrow dy=\frac { x }{ { x }^{ 2 }+1 } dx$

Integrating both sides, we get

$\displaystyle \int { dy } =\int { \frac { x }{ { x }^{ 2 }+1 } } dx\\ \Rightarrow \int { dy } =\frac { 1 }{ 2 } \int { \frac { 2x }{ { x }^{ 2 }+1 } } \\ \Rightarrow y=\frac { 1 }{ 2 } \log { \left| { x }^{ 2 }+1 \right| } +C$

This is the required solution.

Type II  Differential equations of the type $\displaystyle \frac { dy }{ dx } =f\left( y \right)$

To solve this type of differential equations we integrate both sides to obtain the general solution as discussed under:

$\displaystyle \frac { dy }{ dx } =f\left( y \right) \\ \Rightarrow \frac { dx }{ dy } =\frac { 1 }{ f\left( y \right) } \\ \Rightarrow dx=\frac { 1 }{ f\left( y \right) } dy$

Integrating both sides, we obtain

$\displaystyle \int { dx } =\int { \left( { y }^{ 2 }+\sin { y } \right) } dy\\ \Rightarrow x=\frac { { y }^{ 3 } }{ 3 } -\cos { y } +C$

### Example

Solve

$\displaystyle \frac { dy }{ dx } =\frac { 1 }{ { y }^{ 2 }+\sin { y } }$

Solution:

$\displaystyle We\quad have\quad \frac { dy }{ dx } =\frac { 1 }{ { y }^{ 2 }+\sin { y } } \\ \Rightarrow \frac { dx }{ dy } ={ y }^{ 2 }+\sin { y } \\ \Rightarrow dx={ y }^{ 2 }+\sin { y } dy$

Integrating both sides we obtain

$\displaystyle \int { dx } =\int { \left( { y }^{ 2 }+\sin { y } \right) } dy\\ \Rightarrow x=\frac { { y }^{ 3 } }{ 3 } -\cos { y } +C$

This is the required solution.

Type III  Equation in variable separable form

If the differential equation can be put in the form f(x) dx=g(y) dy we say that the variable separable and such equation can be solved by integrating on both sides the equation is given by

$\displaystyle \int { f\left( x \right) } dx=\int { g\left( y \right) } dy+C$

where C is an arbitrary constant.

There is no need of introducing arbitrary constants on both sides as they can be combined together to give just one arbitrary constant.

### Example

Solve

$\displaystyle \left( x+1 \right) \frac { dy }{ dx } =2xy$

Solution:

$\displaystyle We\quad have\quad \left( x+1 \right) \frac { dy }{ dx } =2xy\\ \Rightarrow \left( x+1 \right) dy=2xydx\\ \Rightarrow \frac { dy }{ y } =\frac { 2x }{ x+1 } dx$

Integrating both sides, we get

$\displaystyle \int { \frac { 1 }{ y } } dy=2\int { \frac { x }{ x+1 } } dx\\ \Rightarrow \int { \frac { 1 }{ y } } dy=2\int { \frac { x+1-1 }{ x+1 } } dx\\ \Rightarrow \int { \frac { 1 }{ y } } dy=2\int { \left( 1-\frac { 1 }{ x+1 } \right) } dx\\ \Rightarrow \log { y } =2\left[ x-\log { \left( x+1 \right) } \right] +C$

which is the required solution.

Type IV  Equation reducible to variable separable form

Differential equations of the form $\displaystyle \frac { dy }{ dx } =f\left( ax+by+c \right)$ can be reduced to variable separable form by the substitution ax+by+c=v

### Example

Solve

$\displaystyle \frac { dy }{ dx } ={ \left( 4x+y+1 \right) }^{ 2 }$

Solution:

Given that $\displaystyle \frac { dy }{ dx } ={ \left( 4x+y+1 \right) }^{ 2 }$

Put 4x+y+1=v, so that $\displaystyle 4+\frac { dy }{ dx } =\frac { dv }{ dx } \\ \Rightarrow \frac { dy }{ dx } =\frac { dv }{ dx } -4$

So, the given equation becomes

$\displaystyle \frac { dv }{ dx } -4={ v }^{ 2 }\\ \Rightarrow \frac { dv }{ dx } ={ v }^{ 2 }+4\\ \Rightarrow dv=\left( { v }^{ 2 }+4 \right) dx\\ \Rightarrow \frac { dv }{ { v }^{ 2 }+4 } =dx$

Integrating both sides, we get

$\displaystyle \int { \frac { dv }{ { v }^{ 2 }+4 } } =\int { 1. } dx\\ \Rightarrow \frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { v }{ 2 } \right) } =x+C\\ \Rightarrow \frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { 4x+y+1 }{ 2 } \right) } =x+C$

which is the required solution

Type V   Homogeneous  Differential Equation

A function f(x,y) is called a homogeneous function of degree n if $\displaystyle f\left( \lambda x,\quad \lambda y \right) ={ \lambda }^{ n }f\left( x,y \right)$

For example $\displaystyle f\left( x,y \right) ={ x }^{ 2 }-{ y }^{ 2 }+3xy$ is a homogeneous function degree 2 because $\displaystyle f\left( \lambda x,\quad \lambda y \right) ={ \lambda }^{ 2 }{ x }^{ 2 }-{ \lambda }^{ 2 }{ y }^{ 2 }+3\lambda x.\lambda y={ \lambda }^{ 2 }f\left( x,y \right)$

A homogeneous function f(x,y) of degree n can always be written as

$\displaystyle f\left( x,y \right) ={ x }^{ n }f\left( \frac { y }{ x } \right) \quad or\quad f\left( x,y \right) ={ y }^{ n }f\left( \frac { x }{ y } \right)$

If a first order first degree differential equation is expressible in the form

$\displaystyle \frac { dy }{ dx } =\frac { f\left( x,y \right) }{ g\left( x,y \right) }$

Such type of equations can be reduced to variable separable form by the substitution y=vx as explained below:

The differential equation can be written as

$\displaystyle \frac { dy }{ dx } =\frac { { x }^{ n }f\left( { y }/{ x } \right) }{ { x }^{ n }g\left( { y }/{ x } \right) } \frac { f\left( { y }/{ x } \right) }{ g\left( { y }/{ x } \right) } =F\left( { y }/{ x } \right) \quad \left( say \right)$

If y=vx ,then $\displaystyle \frac { dy }{ dx } =v+x\frac { dy }{ dx }$ Substituting these x values in $\displaystyle \frac { dy }{ dx } =F\left( { y }/{ x } \right)$ we get,

$\displaystyle v+x\frac { dy }{ dx } =F\left( v \right) \\ \Rightarrow \frac { dv }{ F\left( v \right) -v } =\frac { dx }{ x }$

On integration $\displaystyle \int { \frac { dv }{ F\left( v \right) -v } } =\int { \frac { dx }{ x } } +C$

Where C is an arbitrary constant of integration.

After integration v will be replaced by v/x to get the complete solution.

Algorithm for solving homogeneous differential equation

Step I Put the differential equation in the form $\displaystyle \frac { dy }{ dx } =\frac { \phi \left( x,y \right) }{ \psi \left( x,y \right) }$

Step II Put y=vx and $\displaystyle \frac { dy }{ dx } =v+x\frac { dv }{ dx }$ in the equation  in Step I and cancel out x from the right hand side. The equation reduces to form $\displaystyle v+x\frac { dv }{ dx } =f\left( v \right)$

Step III Shift v on RHS and separate the variables v and x.

Step IV Integrate both sides to obtain the solution in terms of v and x.

Step V Replace v by $\displaystyle\frac { y }{ x }$ in the solution obtained in Step IV to obtain the solution in terms of x and y

### Example

Solve the differential  equation $\displaystyle { x }^{ 2 }dy+y\left( x+y \right) dx=0$ given that y=1 when x=1

Solution:

The given differential equation is

$\displaystyle { x }^{ 2 }dy+y\left( x+y \right) dx=0\\ \Rightarrow { x }^{ 2 }dy=-y\left( x+y \right) dx\\ \Rightarrow \frac { dy }{ dx } =-\frac { y\left( x+y \right) }{ { x }^{ 2 } } \\ \Rightarrow \frac { dy }{ dx } =-\left( \frac { xy+{ y }^{ 2 } }{ { x }^{ 2 } } \right) \qquad ........\left( i \right)$

Since each of the functions $\displaystyle xy+{ y }^{ 2 }\quad and\quad { x }^{ 2 }$ is a homogeneous function of degree 2 therefore equation (i) is a homogeneous equation.

Putting y=vx and $\displaystyle \frac { dy }{ dx } =v+x\frac { dy }{ dx }$ in (i) we get

$\displaystyle v+x\frac { dy }{ dx } =-\left( \frac { v{ x }^{ 2 }+{ v }^{ 2 }+{ x }^{ 2 } }{ { x }^{ 2 } } \right) \\ \Rightarrow v+x\frac { dy }{ dx } =-\left( v+{ v }^{ 2 } \right) \\ \Rightarrow x\frac { dv }{ dx } =-2v-{ v }^{ 2 }\\ \Rightarrow xdv=-\left( { v }^{ 2 }+2v \right) dx\\ \Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =-\frac { dx }{ x } \qquad \left[ By\quad separating\quad the\quad variables \right]$

Integrating both sides we get

$\displaystyle \int { \frac { dv }{ { v }^{ 2 }+2v } } =-\int { \frac { dx }{ x } } \\ \Rightarrow \int { \frac { 1 }{ { v }^{ 2 }+2v+1-1 } } dv=-\int { \frac { 1 }{ x } } dx\\ \Rightarrow \int { \frac { 1 }{ { \left( v+1 \right) }^{ 2 }-{ 1 }^{ 2 } } } dv=-\int { \frac { 1 }{ x } } dx$

$\displaystyle \Rightarrow \frac { 1 }{ 2\times 1 } \log { \left( \frac { v+1-1 }{ v+1+1 } \right) } =-\log { x } \log { C } \\ \Rightarrow \frac { 1 }{ 2 } \log { \left( \frac { v }{ v+2 } \right) } =-\log { x } +\log { C } \\ \Rightarrow \log { \left( \frac { v }{ v+2 } \right) } +2\log { x } =2\log { C }$

$\displaystyle \Rightarrow \log { \left( \frac { v }{ v+2 } \right) } +\log { { x }^{ 2 } } =\log { k } \qquad Where\quad k={ C }^{ 2 }\\ \Rightarrow \log { \left( \frac { v }{ v+2 } { x }^{ 2 } \right) } =\log { k } \\ \Rightarrow \frac { v{ x }^{ 2 } }{ v+2 }=k$

$\displaystyle \Rightarrow \frac { \frac { y }{ x } { x }^{ 2 } }{ \frac { y }{ x } +2 } =k\qquad \left[ \because \frac { y }{ x } =v \right] \\ \Rightarrow { x }^{ 2 }y=k\left( y+2x \right) \qquad .....\left( ii \right)$

It is given that y=1, when x=1. Putting x=1 y=1 in eq (ii) we get

$\displaystyle 1=3k\\ \Rightarrow k=\frac { 1 }{ 3 }$

Putting $\displaystyle k=\frac { 1 }{ 3 }$ in (ii), we get  $\displaystyle { x }^{ 2 }y=\left( \frac { 1 }{ 3 } \right) \left( y+2x \right) \\ \Rightarrow 3{ x }^{ 2 }y=\left( y+2x \right)$

This is the required equation.

April 24, 2019