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

Given,

$\frac{dy}{dx}=\frac{x+y-2}{x-y}=\frac{\left(x-1\right)+\left(y-1\right)}{\left(x-1\right)-\left(y-1\right)}$

Now let $x-1=X,y-1=Y$

So, $\frac{dy}{dx}=\frac{X+Y}{X-Y} ........\left(1\right)$

Now let $Y=VX \frac{dY}{dX}=V+X\frac{dV}{dX}$

Putting the value in equation $\left(1\right)$ we get,

$V+X\frac{dV}{dX}=\frac{1+V}{1-V}$

$\Rightarrow X\frac{dV}{dX}=\frac{V^2+1}{1-V}$

$\Rightarrow \int \frac{1-V}{1+V^2}dV=\int \frac{dX}{X}$

$\Rightarrow \int \frac{dV}{1+V^2}-\frac{1}{2}\int \frac{2VdV}{1+V^2}=\int \frac{dX}{X}$

$\Rightarrow {\tan }^{-1}V-\frac{1}{2}\ln \left(1+V^2\right)=\ln X+c$

$\Rightarrow {\tan }^{-1}\left(\frac{Y}{X}\right)-\frac{1}{2}\ln \left(1+\frac{Y^2}{X^2}\right)=\ln \left(X\right)+c$

$\Rightarrow {\tan }^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2}\ln \left(1+\frac{{\left(y-1\right)}^2}{{\left(x-1\right)}^2}\right)=\ln \left(x-1\right)+c$

Now given curve passes through $\left(2,1\right)$

So, $0-\frac{1}{2}\ln 1=\ln 1+c\Rightarrow c=0$

Now given curve also passes through $\left(k+1,2\right)$

So, ${\tan }^{-1}\left(\frac{1}{k}\right)-\frac{1}{2}In\left(1+\frac{1}{{\mathrm{k}}^2}\right)=\ln k$

$\Rightarrow 2{\tan }^{-1}\left(\frac{1}{k}\right)=In\left(\frac{1+k^2}{k^2}\right)+2\ln k$

$\Rightarrow 2{\tan }^{-1}\left(\frac{1}{k}\right)=In\left(1+k^2\right)$