The general solution of the differential equation \( \displaystyle\frac{dy}{dx}-\displaystyle\frac{y}{x}+\mathrm{cosec}\left(\displaystyle\frac{y}{x}\right)=0 \) is
(Consider \( \log x=\log_e x \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \cos\left(\displaystyle\frac{y}{x}\right)=\log x+\mathrm{C} \), where C is an arbitrary constant
The equation involves y and x only through \( \frac{y}{x} \), so it is homogeneous. Let \( y=vx \), so \( \frac{dy}{dx}=v+x\frac{dv}{dx} \):
\( \displaystyle v+x\frac{dv}{dx}-v+\mathrm{cosec}\,v=0 \)
\( \displaystyle x\frac{dv}{dx}=-\mathrm{cosec}\,v \)
The variables separate:
\( \displaystyle \sin v\,dv=-\frac{dx}{x} \)
\( \displaystyle -\cos v=-\log x+C_1 \)
\( \displaystyle \cos v=\log x-C_1 \)
Writing the arbitrary constant \( -C_1 \) as C and putting back \( v=\frac{y}{x} \):
\( \displaystyle \cos\left(\frac{y}{x}\right)=\log x+C \)