If the solution curve, of the differential equation (dy)/(dx)=(x+y-2)/(x-y) passing throug…

Question

If the solution curve, of the differential equation $\frac{dy}{dx}=\frac{x+y-2}{x-y}$ passing through the point $(2,1)$ is ${	an }^{-1}(\frac{y-1}{x-1})-\frac{1}{\beta }{\log }_{e}(\alpha +{(\frac{y-1}{x-1})}^{2})={\log }_{e}|x-1|$, then $5\beta +\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 11

Accepted Answer

$\begin{gathered} \frac{dy}{dx}=\frac{x+y-2}{x-y} \ x=X+h,y=Y+k \ \frac{dY}{dX}=\frac{X+Y}{X-Y} \ \begin{gathered} h+k-2=0 \ h-k=0 \end{gathered}h=k=1 \ Y=vX \ v+\frac{dv}{dX}=\frac{1+v}{1-v}\Rightarrow X-\frac{dv}{dX}=\frac{1+{v}^{2}}{1-v} \end{gathered}$

$\begin{gathered} \frac{1-v}{1+{v}^{2}}dv=\frac{dX}{X} \ {	an }^{-1}v-\frac{1}{2}\ln (1+{v}^{2})=\ln |X|+C \end{gathered}$

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

$\begin{gathered} {	an }^{-1}(\frac{y-1}{x-1})-\frac{1}{2}\ln (1+{(\frac{y-1}{x-1})}^{2})=\ln |x-1| \ 	herefore \alpha =1\beta =2 \ \Rightarrow 5\beta +\alpha =11 \end{gathered}$

Solutions · 1