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

$\frac{d y}{d x}=\frac{x+y-2}{x-y}=\frac{(x-1)+(y-1)}{(x-1)-(y-1)}$ <br/><br/>Let $x-1=X, y-1=Y$ <br/><br/>$\frac{d Y}{d X}=\frac{X+Y}{X-Y}$ <br/><br/>Let $Y=t X \Rightarrow \frac{d Y}{d X}=t+X \frac{d t}{d X}$ <br/><br/>$t+X \frac{d t}{d X}=\frac{1+t}{1-t}$ <br/><br/>$X \frac{d t}{d X}=\frac{1+t}{1-t}-t=\frac{1+t^{2}}{1-t}$ <br/><br/>$\int \frac{1-t}{1+t^{2}} d t=\int \frac{d X}{X}$ <br/><br/>$$ \begin{aligned} &\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^{2}\right)=\ln |X|+c \\\\ &\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+c \end{aligned} $$ <br/><br/>Curve passes through $(2,1)$ <br/><br/>$0-0=0+c \Rightarrow c=0$ <br/><br/>If $(k+1,2)$ also satisfies the curve <br/><br/>$$ \begin{aligned} &\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \ln \left(\frac{1+k^{2}}{k^{2}}\right)=\ln k \\\\ &2 \tan ^{-1}\left(\frac{1}{k}\right)=\ln \left(1+k^{2}\right) \end{aligned} $$