QJEEM24Maf65e: JEE Main 2024 Mathematics - Class 12 JEEM24Maf65e Solution: If the solution curve, of the differential equation passing…

$\begin{aligned}& \frac{d y}{d x}=\frac{x+y-2}{x-y} \\& \mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k} \\& \frac{d Y}{d X}=\frac{X+Y}{X-Y} \\& \left.\begin{array}{l}\mathrm{h}+\mathrm{k}-2=0 \\\mathrm{~h}-\mathrm{k}=0\end{array}\right\} \mathrm{h}=\mathrm{k}=1 \\& \mathrm{Y}=\mathrm{vX} \\& v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^2}{1-v} \\&\end{aligned}$

$\begin{aligned}& \frac{1-v}{1+v^2} d v=\frac{d X}{X} \\& \tan ^{-1} v-\frac{1}{2} \ln \left(1+v^2\right)=\ln |X|+C\end{aligned}$

As curve is passing through $(2,1)$

$\begin{aligned}& \tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln |x-1| \\& \therefore \alpha=1 \text { and } \beta=2 \\& \Rightarrow 5 \beta+\alpha=11\end{aligned}$