Homogeneous Form

Homogeneous form is a differential equation of the first order and first degree in the form \[\frac{dy}{dx}=ƒ\left(\frac{y}{x}\right)\]

A simple test for this is to put $\frac{y}{x}=v$ or $y=vx$, and see whether all $x$’s cancel out or not. If so, it is homogeneous, otherwise not. Hence, putting \[y=vx\] \[\therefore\frac{dy}{dx}=v+x\frac{dv}{dx}\] in the given equation, we get a differential equation with variables $v$ and $x$, i.e. \[v+x\frac{dv}{dx}=ƒ(v)\] \[\therefore\frac{dv}{ƒ(v)-v}=\frac{dx}{x}\] in which the variables are separated. Now, by integrating, we get the required solution. [Variables Separated Form (Standard Form I)]


Solve the following equations:

$\frac{dy}{dx}=\frac{y+x}{x}$

The given equation is obviously homogeneous, since \[\frac{dy}{dx}=\frac{y}{x}+1\] Put $y=vx$, so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=v+1\] \[x\frac{dv}{dx}=1\] \[dv=\frac{dx}{x}\] in which variables are separated.

Now, integrating both sides, we get, \[v=\log x+\log C\] where $C$ is an arbitrary constant. Rewriting $v=\frac{y}{x}$, and simplifying, we get \[y=x\log Cx\] as the required solution.


$xy\frac{dy}{dx}=x^2+y^2$

\[xy\frac{dy}{dx}=x^2+y^2\] \[\frac{dy}{dx}=\frac{x^2+y^2}{xy}\] Put $y=vx$ so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{x^2+v^2x^2}{vx^2}\] \[v+x\frac{dv}{dx}=\frac{1+v^2}{v}\] \[x\frac{dv}{dx}=\frac{1+v^2}{v}-v\] \[x\frac{dv}{dx}=\frac{1}{v}\] \[vdv=\frac{dx}{x}\]

Integrating both sides, we get, \[\frac{v^2}{2}=\log x+C\] \[\frac{y^2}{2x^2}=\log x+C\] \[\therefore y^2=2x^2(\log x+C)\]


$\frac{dy}{dx}=\frac{y}{x}+\tan\frac{y}{x}$

\[\frac{dy}{dx}=\frac{y}{x}+\tan\frac{y}{x}\] Put $y=vx$, so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=v+\tan v\] \[x\frac{dv}{dx}=\tan v\] \[\frac{dv}{\tan v}=\frac{dx}{x}\]

Integrating both sides, we get, \[\log(\sin v)=\log x+\log C\] \[\therefore \sin\frac{y}{x}=Cx\]


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

\[\frac{dy}{dx}=\frac{x^2+y^2}{2x^2}\] \[\frac{dy}{dx}=\frac{1}{2}+\frac{1}{2}\frac{y^2}{x^2}\] Put $y=vx$ so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{1}{2}+\frac{1}{2}v^2\] \[x\frac{dv}{dx}=\frac{1+v^2-2v}{2}\] \[x\frac{dv}{dx}=\frac{(v-1)^2}{2}\] \[\frac{2dv}{(v-1)^2}=\frac{dx}{x}\]

Integrating both sides, we get, \[\frac{-2}{v-1}=\log x+\log C\] \[\frac{-2}{\frac{y}{x}-1}=\log Cx\] \[-2x=(y-x)\log Cx\] \[\therefore 2x=(x-y)\log Cx\]


$xdy-ydx=\sqrt{x^2+y^2}dx$

\[xdy-ydx=\sqrt{x^2+y^2}dx\] \[xdy=ydx+\sqrt{x^2+y^2}dx\] \[\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}\] Put $y=vx$ so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=v+\frac{\sqrt{x^2+v^2x^2}}{x}\] \[v+x\frac{dv}{dx}=\frac{x\sqrt{1+v^2}}{x}\] \[\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}\]

Integrating both sides, we get, \[\log\left(1+\sqrt{1+v^2}\right)=\log x+\log C\] \[\log\left(\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right)=\log Cx\] \[\frac{y}{x}+\sqrt{\frac{x^2+y^2}{x^2}}=Cx\] \[\therefore y+\sqrt{x^2+y^2}=Cx^2\]


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

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

Put $y=vx$ so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{1}{2}(v+v^2)\] \[x\frac{dv}{dx}=\frac{v^2-v}{2}\] \[\frac{2dv}{v^2-v}=\frac{dx}{x}\]

Integrating both sides, we get \[2\log(v-1)-2\log v=\log x+\log C\] \[2\log\left(\frac{y-x}{y}\right)=\log Cx\] \[\therefore (y-x)^2=Cxy^2\]


$(x^3+y^3)dy-x^2ydx=0\text{ }(ƒ(0)=1)$

\[(x^3+y^3)dy-x^ydx=0\] \[\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}\] Put $y=vx$ so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore v+x\frac{dv}{dx}=\frac{vx^3}{x^3+v^3x^3}\] \[v+x\frac{dv}{dx}=\frac{v}{1+v^3}\] \[x\frac{dv}{dx}=\frac{v}{1+v^3}-v\] \[x\frac{dv}{dx}=\frac{v-v-v^4}{1+v^3}\] \[x\frac{dv}{dx}=\frac{-v^4}{1+v^3}\] \[\frac{(1+v^3)}{v^4}dv=-\frac{dx}{x}\] \[\left(-\frac{1}{v^4}-\frac{1}{v}\right)dv=\frac{dx}{x}\]

Integrating both sides, we get, \[\frac{1}{3v^3}-\log v=\log x+\log C\] \[\frac{x^3}{3y^3}=\log\frac{y}{x}+\log x+\log C\] \[x^3=3y^3\log Cy\] When $x=0$, $y=1$, \[\therefore 0=3\log C\] \[\log C=\log 1\] \[\therefore C=1\] \[\therefore x^3=3y^3\log y\]


$\frac{dy}{dx}=\frac{y+1}{y+1+x}$

This equation is not homogeneous. Putting $y+1=z$, then \[\frac{dy}{dx}=\frac{dz}{dx}\] \[\therefore\frac{dz}{dx}=\frac{z}{z+x}=\frac{\left(\frac{z}{x}\right)}{\left(\frac{z}{x}\right)+1}\]

This equation is homogeneous. Put $z=vx$ so that \[\frac{dz}{dx}=v+x\frac{dz}{dx}\] \[v+x\frac{dv}{dx}=\frac{v}{v+1}\] \[x\frac{dv}{dx}=\frac{-v^2}{v+1}\] \[\frac{v+1}{v^2}dv=-\frac{dx}{x}\] \[\frac{dv}{v}+\frac{dv}{v^2}=-\frac{dx}{x}\]

Integrating both sides, we get, \[\log v-v^{-1}=-\log x+\log C\] \[\log v+\log x-\log C=\frac{1}{v}\] \[\log\frac{vx}{C}=\frac{1}{v}\] \[\frac{vx}{C}=e^{\frac{1}{v}}\] \[\frac{z}{C}=e^{\frac{x}{z}}\] \[\therefore y+1=Ce^{\frac{x}{y+1}}\]



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