Question

If the solution curve of the differential equation $\frac{dy}{dx}=\frac{x+y-2}{x-y}$ passes through the points $(2,1)$ and $(k+1,2),k>0$, then

Answer 1

$\frac{dy}{dx}=\frac{x+y-2}{x-y}\dots \left(1\right)$

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

$\frac{dY}{dX}=\frac{X+Y}{X-Y}\dots \left(2\right)$

Where $h+k-2=0$ and $h-k=0$

$\Rightarrow h=k=1$

Equation $\left(2\right)$ is homogenous differential equation so put $Y=vX$ in equation $\left(2\right)$

$v+X\frac{dv}{dX}=\frac{1+v}{1-v}\Rightarrow \frac{v-1}{v^2+1}dv=\frac{-1}{X}dX\dots \left(3\right)$

Integrate

$\ln (v^2+1)-2{\tan }^{-1}v=-2\ln X+c$

$\ln (v^2{X}^2+X^2)=2{\tan }^{-1}v+c$

$\ln (Y^2+X^2)=2{\tan }^{-1}\left(\frac{Y}{X}\right)+c$

$\ln \left[\right(x-1{)}^2+(y-1{)}^2]=2{\tan }^{-1}\left(\frac{y-1}{x-1}\right)+c\dots \left(4\right)$

It is passing through $(2,1)$ so $c=0$

$\Rightarrow \ln \left(\right(x-1{)}^2+(y-1{)}^2)=2{\tan }^{-1}\left(\frac{y-1}{x-1}\right)\dots \left(5\right)$

It is also passing through point $(k+1,2)$ so

$\ln (1+k^2)=2{\tan }^{-1}\left(\frac{1}{k}\right)$