Question

Let \(y=y(x)\) be the solution of the differential equation
\(x\tan\left(\frac{y}{x}\right)dy=\left(y\tan\left(\frac{y}{x}\right)-x\right)dx\)
, \(-1\le x\le1\),
\(y\left(\frac12\right)=\frac{\pi}{6}\)
. Then the area of the region bounded by the curves \(x=0\),
\(x=\frac{1}{\sqrt2}\)
and \(y=y(x)\) in the upper half plane is :

• A. \(\frac{1}{8}(\pi-1)\)
• B. \(\frac{1}{12}(\pi-3)\)
• C. \(\frac{1}{4}(\pi-2)\)
• D. \(\frac{1}{6}(\pi-1)\)

Hint:

Substitute \(y=vx\) to get \(\cos\frac yx=x\), so \(y=x\cos^{-1}x\); then integrate \(x\cos^{-1}x\) from 0 to \(\frac1{\sqrt2}\).

Answer 1

The Correct Option is A

Put \(y=vx\), \(dy=v\,dx+x\,dv\). Then \(x\tan v\,(v\,dx+x\,dv)=(vx\tan v-x)dx\), which reduces to \(x^2\tan v\,dv=-x\,dx\), i.e. \(\tan v\,dv=-\frac{dx}{x}\).
Integrating: \(-\log_e|\cos v|=-\log_e|x|+c\), so \(\cos\frac yx=Cx\).
At \(x=\frac12,\ y=\frac\pi6\): \(\frac yx=\frac\pi3\), \(\cos\frac\pi3=\frac12=C\cdot\frac12\Rightarrow C=1\). So \(\cos\frac yx=x\) and \(y=x\cos^{-1}x\) (which is \(\ge0\) for \(0\le x\le1\), the upper half plane).
Area \(=\int_0^{1/\sqrt2}x\cos^{-1}x\,dx\). By parts: \(\int x\cos^{-1}x\,dx=\frac{x^2}{2}\cos^{-1}x+\frac12\int\frac{x^2}{\sqrt{1-x^2}}dx=\frac{x^2}{2}\cos^{-1}x+\frac14\left(\sin^{-1}x-x\sqrt{1-x^2}\right)\).
At \(x=\frac1{\sqrt2}\): \(\frac14\cdot\frac\pi4+\frac14\left(\frac\pi4-\frac12\right)=\frac{\pi}{8}-\frac18\). At \(x=0\) the value is 0.
Area \(=\frac{1}{8}(\pi-1)\).
Answer: (1) \(\frac{1}{8}(\pi-1)\)