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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

x\sec\left(\dfrac{y}{x}\right)(y\,dx + x\,dy) = y\,\mathrm{cosec}\left(\dfrac{y}{x}\right)(x\,dy - y\,dx)

is:

xy = c\,\mathrm{cosec}\left(\dfrac{y}{x}\right)

xy^2\sin\dfrac{y}{x} = c

xy\,\mathrm{cosec}\dfrac{y}{x} = c

correct

xy = c\sin\left(\dfrac{x}{y}\right)

Solution

Step 1: Rewrite with

\sec = 1/\cos

\mathrm{cosec} = 1/\sin

, and separate the two standard combinations:

\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 = \dfrac{y}{x}

, so

xy = x^2v

y\,dx + x\,dy = d(xy)

, and

x\,dy - y\,dx = x^2\,dv

. Then

\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 \Rightarrow xy = c\sin\frac{y}{x} \Rightarrow xy\,\mathrm{cosec}\frac{y}{x} = c.

Correct answer: (3)

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