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QJEEM22Ma0b96: JEE Main 2022 Mathematics - Class 12 JEEM22Ma0b96 Solution: If the solution curve of the differential equation passes…

If the solution curve of the differential equation dydx=x+y−2x−y\frac{d y}{d x}=\frac{x+y-2}{x-y} passes through the points (2,1)(2,1) and (\mathrm{k}+1,2), \mathrm{k}>0 , then

Multiple Choice Options

  • A 2tan⁡−1(1k)=log⁡e(k2+1)2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)
  • B tan⁡−1(1k)=log⁡e(k2+1)\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)
  • C 2tan⁡−1(1k+1)=log⁡e(k2+2k+2)2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)
  • D 2tan⁡−1(1k)=log⁡e(k2+1k2)2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)

Step-by-Step Verified Solution

dydx=x+y−2x−y=(x−1)+(y−1)(x−1)−(y−1)\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}

Let x−1=X,y−1=Yx-1=X, y-1=Y

dYdX=X+YX−Y\frac{d Y}{d X}=\frac{X+Y}{X-Y}

Let Y=tX⇒dYdX=t+XdtdXY=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}

t+XdtdX=1+t1−tt+X \frac{d t}{d X}=\frac{1+t}{1-t}

XdtdX=1+t1−t−t=1+t21−tX \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}

∫1−t1+t2dt=∫dXX\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}

amp;tan⁡−1t−12ln⁡(1+t2)=ln⁡∣X∣+camp;tan⁡−1(y−1x−1)−12ln⁡(1+(y−1x−1)2)=ln⁡∣x−1∣+c\begin{aligned}&\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\&\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c\end{aligned}

Curve passes through (2,1)(2,1)

0−0=0+c⇒c=00-0=0+c \Rightarrow c=0

If (k+1,2)(k+1,2) also satisfies the curve

amp;tan⁡−1(1k)−12ln⁡(1+k2k2)=ln⁡kamp;2tan⁡−1(1k)=ln⁡(1+k2)\begin{aligned}&\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\&2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right)\end{aligned}

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