Question

Consider the following differential equation:

x(ydx+xdy)cos$\frac{y}{x}=y(xdx-ydx)sin\frac{y}{x}$

Which of the following is the solution of the above equation ( c is an arbitrary constant) ?

• A. $\frac{x}{y}cos\frac{y}{x}=c$
• B. $\frac{x}{y}sin\frac{y}{x}=c$
• C. $xycos\frac{y}{x}=c$
• D. $xysin\frac{y}{x}=c$

Answer 1

The correct option is C $xycos\frac{y}{x}=c$

x(ydx + xdy)cos$\frac{y}{x}=y(xdy-ydx)sin\frac{y}{x}$

$\frac{ydx+xdy}{xdy-ydx}=\frac{y}{x}tan\frac{y}{x}$

Let y = vx

dy = v dx + xdv

$\frac{vxdx+vxdx+x^{2}dy}{vxdx+x^{2}dy-vxdx}=vtanv$

$\frac{xdv+2vdx}{xdv}=vtanv$

1 + $\frac{2v}{x}\frac{dx}{dv}=vtanv$

$\frac{2v}{x}\frac{dx}{dv}=vtanv-1$

$2\frac{dx}{x}=(tanv-\frac{1}{v})dv$

Integrating both sides

2 log x = log $\left|secv\right|$ - log v + log c

$\Rightarrow x^2=\frac{csecv}{v}$

$\Rightarrow x^2.\frac{y}{x}=csec\left(y/x\right)$

$\Rightarrow xycos\frac{y}{x}=c$