Question 25. 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)-\frac1\beta\log_e\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log_e\left|x-1\right|, then 5 \beta+\alpha is equal to
Answer (11)
Explanation:
\frac{d y}{d x}=\frac{x+y-2}{x-y}X = X + h, y = Y + k
\frac{dY}{dX}=\frac{X+Y}{X-Y} \left.\begin{array}{r}h+k-2=0\\h-k=0\end{array}\right\}h=k=1Y = vX
v+\frac{dv}{dX}=\frac{1+v}{1-v}\Rightarrow X-\frac{dv}{dX}=\frac{1+v^2}{1-v} \frac{1 - v}{1 + v^{2}}\, dv = \frac{dX}{X} \tan^{-1} v - \frac{1}{2}\ln\left(1+v^{2}\right) = \ln|X| + CAs curve is passing through (2, 1)
\tan^{-1}\left(\frac{y-1}{x-1}\right)-\frac12\ln\left(1+\left(\frac{y-1}{x-1}\right)^2\right)=\ln\left|x-1\right| \therefore\alpha=1\;and\;\beta=2 \Rightarrow5\beta+\alpha=11