Differential Equations Question 8
Question 8 - 2024 (27 Jan Shift 2)
If the solution curve, of the differential equation $\frac{d y}{d x}=\frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is $\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _e\left(\alpha+\left(\frac{y-1}{x-1}\right)^{2}\right)=\log _e|x-1|$, then $5 \beta+\alpha$ is equal to
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Answer (11)
Solution
$\frac{d y}{d x}=\frac{x+y-2}{x-y}$
$x=X+h, y=Y+k$
$\frac{d Y}{d X}=\frac{X+Y}{X-Y}$
$ \begin{cases} h + k - 2 = 0 \\ h - k = 0 \end{cases} $ $ \quad \Rightarrow \quad h = k = 1 $
$Y=vX$
$v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^{2}}{1-v}$
$\frac{1-v}{1+v^{2}} d v=\frac{d X}{X}$
$\tan ^{-1} v-\frac{1}{2} \ln \left(1+v^{2}\right)=\ln |X|+C$
As curve is passing through $(2,1)$
$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{2} \ln \left(1+\left(\frac{y-1}{x-1}\right)^{2}\right)=\ln |x-1|$
$\therefore \alpha=1$ and $\beta=2$
$\Rightarrow 5 \beta+\alpha=11$
