If the solution curve, of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $5 \beta+\alpha$ is equal to __________.

<p>$$\begin{aligned} & \frac{d y}{d x}=\frac{x+y-2}{x-y} \\ & \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\ & \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\ & \left.\begin{array}{l} \mathrm{h}+\mathrm{k}-2=0 \\ \mathrm{~h}-\mathrm{k}=0 \end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\ & \mathrm{Y}=\mathrm{vX} \\ & v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v} \\ & \end{aligned}$$</p> <p>$$\begin{aligned} & \frac{1-v}{1+v^2} d v=\frac{d X}{X} \\ & \tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C \end{aligned}$$</p> <p>As curve is passing through $(2,1)$</p> <p>$$\begin{aligned} & \tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| \\ & \therefore \alpha=1 \text { and } \beta=2 \\ & \Rightarrow 5 \beta+\alpha=11 \end{aligned}$$</p>