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

Question

$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$

$\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$

$2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)$

$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)$

Solution

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