Question

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

• A. $xycos\left(\frac{y}{x}\right)=K$
• 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)

$xycos\left(\frac{y}{x}\right)=K$

$y[x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)]dx=x[y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)]dy$

$xy\cos \frac{y}{x}[1+\frac{y}{x}\tan \frac{y}{x}]dx=x^2\cos \frac{y}{x}[\frac{y}{x}\tan \frac{y}{x}-1]dy$

$\frac{dy}{dx}=-\frac{\frac{y}{x}[1+\frac{y}{x}\tan \frac{y}{x}]}{[1-\frac{y}{x}\tan \frac{y}{x}]}$

Let $\frac{y}{x}=t$

$\frac{dy}{dx}=t+x\frac{dt}{dx}$

$t+x\frac{dt}{dx}=-t\frac{[1+t\tan t]}{[1-t\tan t]}$

$x\frac{dt}{dx}=\frac{-2t}{1-t\tan t}$

$\frac{t\tan t-1}{2t}dt=\frac{dx}{x}$

Upon integration ,

$\int \tan tdt-\int \frac{1}{t}dt=2\int \frac{dx}{x}$

$-\ln \cos t=\ln \left(x^{2}t\right)-\ln K$

$-\ln \left(\cos t\right)x^{2}t=-\ln K$

$xy\cos \left(\frac{y}{x}\right)=K$