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
$\frac{d y}{d x}=\frac{x+y-2}{x-y}$
$\mathrm{x}=\mathrm{X}+\mathrm{h}, \mathrm{y}=\mathrm{Y}+\mathrm{k}$
$\frac{d Y}{d X}=\frac{X+Y}{X-Y}$
$\left.\begin{array}{l}\mathrm{h}+\mathrm{k}-2=0 \ \mathrm{~h}-\mathrm{k}=0\end{array}\right} \mathrm{h}=\mathrm{k}=1$
$\mathrm{Y}=\mathrm{vX}$
$v+\frac{d v}{d X}=\frac{1+v}{1-v} \Rightarrow X-\frac{d v}{d X}=\frac{1+v^{2}}{1-v}$
$\frac{1-v}{1+v^{2}} d v=\frac{d X}{X}$
$\tan ^{-1} v-\frac{1}{2} \ln \left(1+v^{2}\right)=\ln |X|+C$
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 $\beta=2$
$\Rightarrow 5 \beta+\alpha=11$
