QJEEM21Maf681: JEE Main 2021 Mathematics - Class 12 JEEM21Maf681 Solution: Let y = y(x) be the solution of the differential equation , , . …

Question

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Let y = y(x) be the solution of the differential equation

x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx

,

- 1 \le x \le 1

,

y\left( {{1 \over 2}} \right) = {\pi \over 6}

. Then the area of the region bounded by the curves x = 0,

x = {1 \over {\sqrt 2 }}

and y = y(x) in the upper half plane is :

Accepted Answer

We have,

{{dy} \over {dx}} = {{x\left( {{y \over x}.\tan {y \over x} - 1} \right)} \over {x\tan {y \over x}}}

\therefore

{{dy} \over {dx}} = {y \over x} - \cot \left( {{y \over x}} \right)

Put

{y \over x} = v

\Rightarrow y = vn

\therefore

{{dy} \over {dx}} = v + {{ndv} \over {dx}}

Now, we get

v + n{{dv} \over {dx}} = v - \cot (v)

\Rightarrow \int {(\tan )dv} = - \int {{{dx} \over x}}

\therefore

\ln \left| {\sec \left( {{y \over x}} \right)} \right| = - \ln \left| x \right| + c

As,

\left( {{1 \over 2}} \right) = \left( {{y \over x}} \right) \Rightarrow C = 0

\therefore

\sec \left( {{y \over x}} \right) = {1 \over x}

\Rightarrow \cos \left( {{y \over x}} \right) = x

\therefore

y = x{\cos ^{ - 1}}(x)

So, required bounded area

= \int\limits_0^{{1 \over {\sqrt 2 }}} {\mathop x\limits_{(II)} (\mathop {{{\cos }^{ - 1}}x}\limits_{(I)} )dx = \left( {{{\pi - 1} \over 8}} \right)}

(I. B. P.)

\therefore

Option (1) is correct.

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