Question

Solution of the differential equation
$y\left[x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right] d x$
$-x\left[y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right] d y=0$

• A. $ xy\cos \left( \frac{y}{x} \right)=K $
• B. $ \cos \left( \frac{x}{y} \right)=K\,xy $
• C. $ \frac{x}{y}\cos \left( \frac{y}{x} \right)=K $
• D. None of these

Answer 1

Answer: (A)

The given differential equation is
$y\left[x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right] d x$
$-x\left[y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right] d y=0$
$\Rightarrow \frac{d y}{d x}=\frac{y\left[x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right]}{x\left[y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right]}$
Putting $y=v x$,
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ in Eq... (i), we get
$v+x \frac{d v}{d x}=\frac{v x[x \cos v+v x \sin v]}{x[v x \sin v-x \cos v]}$
$\Rightarrow x \frac{d v}{d x}=\frac{v x \cos v+v^{2} x \sin v}{v x \sin v-x \cos v}-v$
$\Rightarrow x \frac{d v}{d x}$
$=\frac{v x \operatorname{cosv}+v^{2} x \sin v-v^{2} x \sin v+x v \cos v}{v x \sin v-x \cos v}$
$\Rightarrow x \frac{d v}{d x}=\frac{2 x v \cos v}{x v \sin v-x \cos v}$
$\Rightarrow \frac{v(\sin v-\cos v) d v}{v \cos v}=2 \frac{d x}{x}$
Integrating both sides, we get
$-\int \frac{(\cos v-v \sin v) d v}{v \cos v}=2 \int \frac{d x}{x}$
$\Rightarrow -\log |v \cos v|=2 \log |x|+\log c$
$\Rightarrow \log \left|\frac{1}{v \cos v}\right|=\log \left|x^{2}\right|+\log c$
$\Rightarrow \frac{1}{v \cos v}=c x^{2}$
$\Rightarrow \frac{x}{y} \sec \left(\frac{y}{x}\right)=c x^{2}$
$\Rightarrow x y \cos \left(\frac{y}{x}\right)=\frac{1}{c}$
$\Rightarrow x y \cos \left(\frac{y}{x}\right)=k,\left(k=\frac{1}{c}\right)$