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Homogeneous Differential Equation with sec and cosec | JEE

JEE Maths question with a full step-by-step solution.

Question
The solution of xsec⁡(yx)(y dx+x dy)=y cosec(yx)(x dy−y dx)x\sec\left(\dfrac{y}{x}\right)(y\,dx + x\,dy) = y\,\mathrm{cosec}\left(\dfrac{y}{x}\right)(x\,dy - y\,dx) is:
Axy=c cosec(yx)xy = c\,\mathrm{cosec}\left(\dfrac{y}{x}\right)
Bxy2sin⁡yx=cxy^2\sin\dfrac{y}{x} = c
Cxy cosecyx=cxy\,\mathrm{cosec}\dfrac{y}{x} = ccorrect
Dxy=csin⁡(xy)xy = c\sin\left(\dfrac{x}{y}\right)
Solution
Step 1: Rewrite with sec⁡=1/cos⁡\sec = 1/\cos, cosec=1/sin⁡\mathrm{cosec} = 1/\sin, and separate the two standard combinations:
y dx+x dyx dy−y dx=ycos⁡(yx)xsin⁡(yx)=yxcot⁡yx.\frac{y\,dx + x\,dy}{x\,dy - y\,dx} = \frac{y\cos\left(\frac{y}{x}\right)}{x\sin\left(\frac{y}{x}\right)} = \frac{y}{x}\cot\frac{y}{x}.
 Step 2: Put v=yxv = \dfrac{y}{x}, so xy=x2vxy = x^2v, y dx+x dy=d(xy)y\,dx + x\,dy = d(xy), and x dy−y dx=x2 dvx\,dy - y\,dx = x^2\,dv. Then
d(xy)x2 dv=vcot⁡v⇒d(xy)xy=vcot⁡v x2 dvx2v=cot⁡v dv.\frac{d(xy)}{x^2\,dv} = v\cot v \Rightarrow \frac{d(xy)}{xy} = \frac{v\cot v\, x^2\,dv}{x^2 v} = \cot v\,dv.
 Step 3: Integrate:
ln⁡(xy)=ln⁡sin⁡v+ln⁡c⇒xy=csin⁡yx⇒xy cosecyx=c.\ln(xy) = \ln\sin v + \ln c \Rightarrow xy = c\sin\frac{y}{x} \Rightarrow xy\,\mathrm{cosec}\frac{y}{x} = c.
 Correct answer: (3)
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