उत्तर (11)

समाधान

$\dfrac{d y}{d x}=\dfrac{x+y-2}{x-y}$

$x=X+h, y=Y+k$

$\dfrac{d Y}{d X}=\dfrac{X+Y}{X-Y}$

$ \begin{cases} h + k - 2 = 0 \\ h - k = 0 \end{cases} $ $ \quad \Rightarrow \quad h = k = 1 $

$Y=vX$

$v+X\dfrac{d v}{d X}=\dfrac{1+v}{1-v} \Rightarrow X\dfrac{d v}{d X}=\dfrac{1+v^{2}}{1-v}$

$\dfrac{1-v}{1+v^{2}} d v=\dfrac{d X}{X}$

$\tan ^{-1} v-\dfrac{1}{2} \ln \left(1+v^{2}\right)=\ln |X|+C$

$\tan ^{-1}\left(\dfrac{y-1}{x-1}\right)-\dfrac{1}{2} \ln \left(1+\left(\dfrac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|+C $

जबकि वक्र $(2,1)$ से गुजरता है, हमें प्राप्त होता है

$ C=0 $

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

$\therefore \alpha=1$ और $\beta=2$

$\Rightarrow 5 \beta+\alpha=11$