Differential Equations

# 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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