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
\[If,f\left( tx,ty \right)={{t}^{n}}f(x,y),t>0\]

\[f\left( x,y \right)=\frac{{{x}^{5}}+{{y}^{5}}}{{{x}^{2}}-{{y}^{2}}}\]
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

\[Solve:\frac{dy}{dx}=\frac{y}{x}+\sin \frac{y}{x}\]
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.

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

\[Solve:2xydx-\left( {{x}^{2}}-{{y}^{2}} \right)dy=0\]
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.

\[Solve:\left( 1+{{e}^{\frac{x}{y}}} \right)dx+{{e}^{\frac{y}{x}}}\left( 1-\frac{x}{y} \right)dy=0\]
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.

\[Solve:{{x}^{3}}\frac{dy}{dx}={{y}^{3}}+{{y}^{2}}\sqrt{{{y}^{2}}-{{x}^{2}}}\]
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.

\[Solve:x\cos \left( \frac{y}{x} \right)\left( ydx+xdy \right)=y\sin \left( \frac{y}{x} \right)\left( xdy-ydx \right)\]
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.

\[Solve:\frac{dy}{dx}=\frac{x+y+1}{3x+3y+1}\]
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.

\[Solve:\left( 3x+2y-5 \right)dx+\left( 2x+3y-5 \right)dy=0\]
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.;