Question

Show that the given differential equation is homogeneous and then solve it.

$\{xcos\left(\frac{y}{x}\right)+ysin\left(\frac{y}{x}\right)\}ydx=\{ysin\left(\frac{y}{x}\right)-xcos\left(\frac{y}{x}\right)\}xdy$

Answer 1

Given, $\{xcos\left(\frac{y}{x}\right)+ysin\left(\frac{y}{x}\right)\}ydx=\{ysin\left(\frac{y}{x}\right)-xcos\left(\frac{y}{x}\right)\}xdy$
$\Rightarrow \frac{dy}{dx}=\frac{y\{xcos\frac{y}{x}+ysin\frac{y}{x}\}}{x\{ysin\frac{y}{x}-xcos\frac{y}{x}\}}...\left(i\right)$
Thus, the given differential equations is homogeneous.
So, put y=vx
$\frac{dy}{dx}=v+x\frac{dv}{dx}\therefore v+x\frac{dv}{dx}=\frac{vx(xcosv+vxsinv)}{x(vxsinv-xcosv)}\Rightarrow v+x\frac{dv}{dx}=\frac{v(cosv+vsinv)}{vsinv-cosv}\Rightarrow x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv}{vsinv-cosv}-v\Rightarrow x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv-v^{2}sinv+vcosv}{vsinv-cosv}\Rightarrow x\frac{dv}{dx}=\frac{2vcosv}{vsinv-cosv}\Rightarrow \left(\frac{vsinv-cosv}{vcosv}\right)dv=\frac{2}{x}dx\Rightarrow (tanv-\frac{1}{v})dv=\frac{2}{x}dx$
On integrating both sides, we get $\int (tanv-\frac{1}{v})dv=\int \frac{2dx}{x}$
$\Rightarrow \int tanvdv-\int \frac{1}{v}dv=2\int \frac{1}{x}dx\Rightarrow -log|cosv|-log|v|=2log|x|+C\Rightarrow log|vcosv|+2log|x|=-C$ $(\because logm+logn=logmn)$
$log\left[\right(vcosv\left)x^2\right]=-C\Rightarrow \left(vcosv\right)x^2=e^{-c}$ $\because lo{g}_{e}x=m\Rightarrow e^m=x)$
$\Rightarrow x^{2}vcosv=A$ $(where,A=e^{-c})$
$\Rightarrow x^2\frac{y}{x}cos\frac{y}{x}=A\Rightarrow xycos\frac{y}{x}=A$
This is the required solution of the given differential equation.