Question

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

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

Answer 1

Answer: (A)

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