# Must Do Problems Of Homogeneous Differential Equation

Differential Equation / Sunday, September 2nd, 2018

# Homogeneous differential equation

Before starting to solve our must do problems of homogeneous differential equation let us first understand, what is Homogeneous function?

## Homogeneous Function

A function f(x, y) of two variables x, y is said to be a homogeneous function of degree n, if f(x, y) can be expressed as either
${{x}^{n}}\phi \left( \frac{y}{x} \right),or,{{y}^{n}}\psi \left( \frac{x}{y} \right)$
OR
$If,f\left( tx,ty \right)={{t}^{n}}f(x,y),t>0$

 Example

$f\left( x,y \right)=\frac{{{x}^{5}}+{{y}^{5}}}{{{x}^{2}}-{{y}^{2}}}$
Is a homogeneous function of x and y of degree 3. As
$f\left( x,y \right)=\frac{{{x}^{5}}+{{y}^{5}}}{{{x}^{2}}-{{y}^{2}}}=\frac{{{x}^{5}}\left\{ 1+{{\left( \frac{y}{x} \right)}^{5}} \right\}}{{{x}^{2}}\left\{ 1-{{\left( \frac{y}{x} \right)}^{2}} \right\}}={{x}^{3}}\phi \left( \frac{y}{x} \right)$
Also
$f\left( tx,ty \right)=\frac{{{t}^{5}}{{x}^{5}}+{{t}^{5}}{{y}^{5}}}{{{t}^{2}}{{x}^{2}}-{{t}^{2}}{{y}^{2}}}={{t}^{3}}.\frac{{{x}^{5}}+{{y}^{5}}}{{{x}^{2}}-{{y}^{2}}}={{t}^{3}}f(x,y)$

Note:
Before Continuing I Suggest you to first go through my previous articles on
01. Introduction To Differential Equation – What? Why? How?
02. Cheat Sheet Of First order first degree differential equation Solution Methods
03. 21 Most Valuable Questions to Master – Separation of Variables Method

 Question 01

$Solve:\frac{dy}{dx}=\frac{y}{x}+\sin \frac{y}{x}$
Solution:
$Put,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
The given equation becomes,
$v+x\frac{dv}{dx}=v+\sin v$
$\Rightarrow x\frac{dv}{dx}=\sin v\Rightarrow \cos ecvdv=\frac{dx}{x}$
Integrating both sides we get,
$\int{\cos ecvdv}=\int{\frac{dx}{x}+\log c}$
$\Rightarrow \log \left( {{\tan }^{-1}}\frac{v}{2} \right)=\log x+\log c$
$\Rightarrow \log \left( {{\tan }^{-1}}\frac{v}{2} \right)=\log cx$
$\Rightarrow {{\tan }^{-1}}\frac{v}{2}=cx\Rightarrow \frac{v}{2}=\tan cx$
$\Rightarrow \frac{y}{2x}=\tan cx\Rightarrow y=2x\tan cx$
c being arbitrary constant.

 Question 02

$Solve:\frac{dy}{dx}+\frac{y}{x}=\frac{{{y}^{2}}}{{{x}^{2}}}$
Solution:
$Put,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
The given equation becomes,
$v+x\frac{dv}{dx}+v={{v}^{2}}$
$\Rightarrow x\frac{dv}{dx}={{v}^{2}}-2v$
$\Rightarrow \frac{dv}{{{v}^{2}}-2v}=\frac{dx}{x}$
Integrating both sides we get,
$\Rightarrow \int{\frac{dv}{{{v}^{2}}-2v}}=\int{\frac{dx}{x}+\log c}$
$\Rightarrow \int{\frac{dv}{{{v}^{2}}-2v+1-1}}=\int{\frac{dx}{x}+\log c}$
$\Rightarrow \int{\frac{dv}{{{\left( v-1 \right)}^{2}}-1}}=\int{\frac{dx}{x}+\log c}$
$\Rightarrow \frac{1}{2.1}\log \left| \frac{v-1-1}{v-1+1} \right|=\log x+\log c$
$\Rightarrow \frac{1}{2}\log \left| \frac{v-2}{v} \right|=\log cx$
$\Rightarrow \log \left| \frac{v-2}{v} \right|=\log {{\left( cx \right)}^{2}}$
$\Rightarrow \frac{\frac{y}{x}-2}{\frac{y}{x}}={{c}^{2}}{{x}^{2}}\Rightarrow \frac{y-2x}{y}={{c}^{2}}{{x}^{2}}$
$\therefore y-2x={{c}^{2}}{{x}^{2}}y$
c being arbitrary constant.

 Question 03

$Solve:2xydx-\left( {{x}^{2}}-{{y}^{2}} \right)dy=0$
Solution:
$Here,\frac{dy}{dx}=\frac{2xy}{\left( {{x}^{2}}-{{y}^{2}} \right)}=\frac{2\left( \frac{y}{x} \right)}{1-{{\left( \frac{y}{x} \right)}^{2}}}$
This is a homogeneous equation.
$Put,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
The given equation becomes,
$v+x\frac{dv}{dx}=\frac{2v}{1-{{v}^{2}}}$
$\Rightarrow x\frac{dv}{dx}=\frac{2v}{1-{{v}^{2}}}-v=\frac{2v-v+{{v}^{3}}}{1-{{v}^{2}}}$
$\Rightarrow x\frac{dv}{dx}=\frac{v+{{v}^{3}}}{1-{{v}^{2}}}$
$\Rightarrow \frac{1-{{v}^{2}}}{v+{{v}^{3}}}dv=\frac{dx}{x}$
$\Rightarrow \frac{1+{{v}^{2}}-2{{v}^{2}}}{v\left( 1+{{v}^{2}} \right)}=\frac{dx}{x}$
$\Rightarrow \left[ \frac{1}{v}-\frac{2v}{1+{{v}^{2}}} \right]dv=\frac{dx}{x}$
Integrating both sides we get,
$\int{\left[ \frac{1}{v}-\frac{2v}{1+{{v}^{2}}} \right]dv}=\int{\frac{dx}{x}+\log c}$
$\Rightarrow \log v-\log \left( 1+{{v}^{2}} \right)=\log x+\log c$
$\Rightarrow \log \left( \frac{v}{1+{{v}^{2}}} \right)=\log cx$
$\Rightarrow cx=\frac{v}{1+{{v}^{2}}}=\frac{\frac{y}{x}}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}=\frac{xy}{{{x}^{2}}+{{y}^{2}}}$
$\therefore y=c\left( {{x}^{2}}+{{y}^{2}} \right)$
$\therefore y=c\left( {{x}^{2}}+{{y}^{2}} \right)$
c being arbitrary constant.

 Question 04

$Solve:\left( 1+{{e}^{\frac{x}{y}}} \right)dx+{{e}^{\frac{y}{x}}}\left( 1-\frac{x}{y} \right)dy=0$
Solution:
$Given,\left( 1+{{e}^{\frac{x}{y}}} \right)dx+{{e}^{\frac{y}{x}}}\left( 1-\frac{x}{y} \right)dy=0$
$\Rightarrow \left( 1+{{e}^{\frac{x}{y}}} \right)dx={{e}^{\frac{y}{x}}}\left( \frac{x}{y}-1 \right)dy$
$\Rightarrow \frac{dx}{dy}=\frac{{{e}^{\frac{y}{x}}}\left( \frac{x}{y}-1 \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}$
This is a homogeneous equation.
$Put,x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$
The given equation becomes,
$v+y\frac{dv}{dy}=\frac{{{e}^{v}}\left( v-1 \right)}{\left( 1+{{e}^{v}} \right)}$
$\Rightarrow y\frac{dv}{dy}=\frac{{{e}^{v}}\left( v-1 \right)}{\left( 1+{{e}^{v}} \right)}-v=\frac{v{{e}^{v}}-{{e}^{v}}-v-v{{e}^{v}}}{\left( 1+{{e}^{v}} \right)}$
$\Rightarrow y\frac{dv}{dy}=-\frac{\left( v+{{e}^{v}} \right)}{\left( 1+{{e}^{v}} \right)}$
$\Rightarrow \frac{1+{{e}^{v}}}{v+{{e}^{v}}}dv=-\frac{dy}{y}$
Integrating both sides we get,
$\int{\frac{1+{{e}^{v}}}{v+{{e}^{v}}}dv}=-\int{\frac{dy}{y}+\log c}$
$\Rightarrow \log \left( v+{{e}^{v}} \right)+\log y=\log c$
$\Rightarrow \log y\left( v+{{e}^{v}} \right)=\log c$
$\Rightarrow y\left( \frac{x}{y}+{{e}^{\frac{x}{y}}} \right)=c$
$\therefore x+y{{e}^{\frac{x}{y}}}=c$
c being arbitrary constant.

 Question 05

$Solve:{{x}^{3}}\frac{dy}{dx}={{y}^{3}}+{{y}^{2}}\sqrt{{{y}^{2}}-{{x}^{2}}}$
Solution:
$We-have,\frac{dx}{dy}=\frac{{{x}^{3}}}{{{y}^{3}}+{{y}^{2}}\sqrt{{{y}^{2}}-{{x}^{2}}}}$
$Put,x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$
The given equation becomes,
$v+y\frac{dv}{dy}=\frac{{{v}^{3}}{{y}^{3}}}{{{y}^{3}}+{{y}^{2}}\sqrt{{{y}^{2}}-{{v}^{2}}{{y}^{2}}}}$
$\Rightarrow y\frac{dv}{dy}=\frac{{{v}^{3}}}{1+\sqrt{1-{{v}^{2}}}}-v$
$\Rightarrow y\frac{dv}{dy}=\frac{{{v}^{3}}-v-v\sqrt{1-{{v}^{2}}}}{1+\sqrt{1-{{v}^{2}}}}$
$\Rightarrow y\frac{dv}{dy}=\frac{-v\sqrt{1-{{v}^{2}}}\left( 1+\sqrt{1-{{v}^{2}}} \right)}{1+\sqrt{1-{{v}^{2}}}}$
$\Rightarrow y\frac{dv}{dy}=-v\sqrt{1-{{v}^{2}}}$
$\Rightarrow \frac{dv}{v\sqrt{1-{{v}^{2}}}}+\frac{dy}{y}=0$
$Let,v=\frac{1}{z}\Rightarrow dv=-\frac{1}{{{z}^{2}}}dz$
$\Rightarrow \frac{-\frac{1}{{{z}^{2}}}dz}{\frac{1}{z}\sqrt{1-\frac{1}{{{z}^{2}}}}}+\frac{dy}{y}=0$
$\Rightarrow \frac{dz}{\sqrt{{{z}^{2}}-1}}-\frac{dy}{y}=0$
Integrating we get,
$\int{\frac{dz}{\sqrt{{{z}^{2}}-1}}}-\int{\frac{dy}{y}}=\log c$
$\Rightarrow \log \left( z+\sqrt{{{z}^{2}}-1} \right)=\log c+\log y$
$\Rightarrow z+\sqrt{{{z}^{2}}-1}=cy$
$\Rightarrow \frac{1}{v}+\sqrt{\frac{1}{{{v}^{2}}}-1}=cy$
$\Rightarrow 1+\sqrt{1-{{v}^{2}}}=cvy$
$\Rightarrow 1+\sqrt{1-\frac{{{x}^{2}}}{{{y}^{2}}}}=c.\frac{x}{y}.y$
$\therefore y+\sqrt{{{y}^{2}}-{{x}^{2}}}=cxy$
c being arbitrary constant.

 Question 06

$Solve:x\cos \left( \frac{y}{x} \right)\left( ydx+xdy \right)=y\sin \left( \frac{y}{x} \right)\left( xdy-ydx \right)$
Solution:
$Given,x\cos \left( \frac{y}{x} \right)\left( ydx+xdy \right)=y\sin \left( \frac{y}{x} \right)\left( xdy-ydx \right)$
$\Rightarrow x\cos \left( \frac{y}{x} \right).ydx+x\cos \left( \frac{y}{x} \right).xdy=y\sin \left( \frac{y}{x} \right).xdy-y\sin \left( \frac{y}{x} \right).ydx$
$\Rightarrow x\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}dy=y\left\{ y\sin \left( \frac{y}{x} \right)+x\cos \left( \frac{y}{x} \right) \right\}dx$
$\Rightarrow \frac{dy}{dx}=\frac{y\left\{ y\sin \left( \frac{y}{x} \right)+x\cos \left( \frac{y}{x} \right) \right\}}{x\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}}$
This is a homogeneous equation.
$Put,y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
The given equation becomes,
$v+x\frac{dv}{dx}=\frac{v\left( v\sin v+\cos v \right)}{v\sin v-\cos v}$
$\Rightarrow x\frac{dv}{dx}=\frac{v\left( v\sin v+\cos v \right)}{v\sin v-\cos v}-v=\frac{2v\cos v}{v\sin v-\cos v}$
$\Rightarrow \frac{2}{x}dx=\frac{v\sin v-\cos v}{v\cos v}dv$
$\Rightarrow \frac{2}{x}dx=\left( \tan v-\frac{1}{v} \right)dv$
Integrating we get,
$2\int{\frac{1}{x}dx}=\int{\left( \tan v-\frac{1}{v} \right)dv+\log c}$
$\Rightarrow 2\log x=\log scev-\log v+\log c$
$\Rightarrow \log {{x}^{2}}=\log \left( \frac{c\sec v}{v} \right)$
$\Rightarrow {{x}^{2}}=\frac{c\sec \frac{y}{x}}{\frac{y}{x}}$
$\therefore \sec \frac{y}{x}=cxy$
c being arbitrary constant.

## Non-homogeneous equations reducible to homogeneous form

Non-homogeneous equations of certain forms can be made homogeneous by simple substitution. We shall discuss here the method of solution of differential equation of the form
$\frac{dy}{dx}=\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}……….(1)$
Two different cases will be considered.
$Type-1:-Here,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=k(\ne 0)$
In this case ${{a}_{2}}x+{{b}_{2}}y=z$will be the proper substitution.
$Then,\frac{dz}{dx}={{a}_{2}}+{{b}_{2}}\frac{dy}{dx}$
and the equation (1) becomes
$\frac{dy}{dx}=\frac{k\left( {{a}_{2}}x+{{b}_{2}}y \right)+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}$
$\Rightarrow \frac{dz}{dx}={{a}_{2}}+{{b}_{2}}\frac{kz+{{c}_{1}}}{z+{{c}_{2}}}$
Where the variables may be separated and the required general solution will be obtained as usual.
$Type-2:-Here,\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$
Here, a change of variables will transform the given differential equation to a homogeneous one.

We substitute, $x=X+h,y=Y+k$where X and Y are new variables and h, k are constants, chosen such a way that,
${{a}_{1}}h+{{b}_{1}}k+{{c}_{1}}=0$
${{a}_{2}}h+{{b}_{2}}k+{{c}_{2}}=0$
With these substitutions, equation (1) reduces to
$\frac{dY}{dX}=\frac{{{a}_{1}}X+{{b}_{1}}Y}{{{a}_{2}}X+{{b}_{2}}Y}$
Which, being a homogeneous equation, can be solved by methods discussed in this article.

 Question 07

$Solve:\frac{dy}{dx}=\frac{x+y+1}{3x+3y+1}$
Solution:
$Given,\frac{dy}{dx}=\frac{x+y+1}{3x+3y+1}$
is of the form,
$\frac{dy}{dx}=\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}},where,\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{3}$
$So-put,x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}$
Therefor the given equation becomes
$\frac{dz}{dx}-1=\frac{z+1}{3z+1}$
$\Rightarrow \frac{dz}{dx}=\frac{z+1}{3z+1}+1=\frac{4z+2}{3z+1}$
$\Rightarrow \frac{3z+1}{2z+1}dz=2dx$
$\Rightarrow 2dx=\left\{ \frac{3}{2}-\frac{1}{2\left( 2z+1 \right)} \right\}dz$
Integrating we get,
$\int{2dx}=\int{\left\{ \frac{3}{2}-\frac{1}{2\left( 2z+1 \right)} \right\}dz+k}$
$\Rightarrow 2x=\frac{3}{2}z-\frac{1}{4}\log \left( 2z+1 \right)+k$
$\therefore 2x=\frac{3}{2}\left( x+y \right)-\frac{1}{4}\log \left( 2x+2y+1 \right)+k$
k being arbitrary constant.

 Question 08

$Solve:\left( 3x+2y-5 \right)dx+\left( 2x+3y-5 \right)dy=0$
Solution:
$Here,\frac{dy}{dx}=-\frac{3x+2y-5}{2x+3y-5}$
$Let-put,x=X+h,y=Y+k,$
Where h, k are constants.
$Now,\frac{dy}{dx}=\frac{dY}{dX},equation-becomes$
$\frac{dy}{dx}=-\frac{3\left( X+h \right)+2\left( Y+k \right)-5}{2\left( X+h \right)+3\left( Y+k \right)-5}$
$\Rightarrow \frac{dy}{dx}=-\frac{3X+2Y+3h+2k-5}{2X+3Y+2h+3k-5}$
H, k are chosen in such a way that
$3h+2k-5=0$
$2h+3k-5=0$
$i.e.,\frac{h}{-10+15}=\frac{k}{-10+15}=\frac{1}{9-4}$
$\therefore h=1,k=1$
$\therefore \frac{dy}{dx}=-\frac{3X+2Y}{2X+3Y}$
which is homogeneous.
$Put,Y=vX\Rightarrow \frac{dY}{dX}=v+X\frac{dv}{dX}$
And the above equation becomes,
$v+X\frac{dv}{dX}=-\frac{3+2v}{2+3v}$
$\Rightarrow X\frac{dv}{dX}=-\frac{3+2v}{2+3v}-v=-\left\{ \frac{3{{v}^{2}}+4v+3}{2+3v} \right\}$
$\Rightarrow \frac{2+3v}{3{{v}^{2}}+4v+3}dv=-\frac{dX}{X}$
Integrating we get
$\int{\frac{2+3v}{3{{v}^{2}}+4v+3}dv}=-\int{\frac{dX}{X}+\log c}$
$\Rightarrow \frac{1}{2}\log \left| 3{{v}^{2}}+4v+3 \right|=-\log X+\log c$
$\Rightarrow \log \left| 3{{v}^{2}}+4v+3 \right|=\log \frac{{{c}^{2}}}{{{X}^{2}}}$
$\Rightarrow 3{{v}^{2}}+4v+3=\frac{{{c}^{2}}}{{{X}^{2}}}$
$\Rightarrow 3{{v}^{2}}{{X}^{2}}+4v{{X}^{2}}+3{{X}^{2}}={{c}^{2}}$
$\Rightarrow 3{{Y}^{2}}+4XY+3{{X}^{2}}={{c}^{2}}$
$\Rightarrow 3{{\left( y-1 \right)}^{2}}+4\left( x-1 \right)\left( y-1 \right)+3{{\left( x-1 \right)}^{2}}={{c}^{2}}$
$\therefore 3\left( {{x}^{2}}+{{y}^{2}} \right)-10\left( x+y \right)+4xy=k,\left[ where,k={{c}^{2}}-10 \right]$
is the required general solution.