1. Given differential equation

$\frac{dy}{dx}=\frac{x+y-2}{x-y}$

We are told that the same solution curve passes through:

$(2,1) \quad \text{and} \quad (k+1,2), \; k>0$

We need the relation satisfied by $k$.

2. Shift the origin to simplify

Observe the expression has $x+y-2$. Let us put

$X=x-1, \qquad Y=y-1$

Then

$x=X+1, \qquad y=Y+1$

and since shifting does not change derivative,

$\frac{dy}{dx}=\frac{dY}{dX}$

Now,

$x+y-2=(X+1)+(Y+1)-2=X+Y$

and

$x-y=(X+1)-(Y+1)=X-Y$

So the differential equation becomes

$\frac{dY}{dX}=\frac{X+Y}{X-Y}$

This is a homogeneous differential equation.

3. Use substitution $Y=vX$

Let

$Y=vX$

Then

$\frac{dY}{dX}=v+X\frac{dv}{dX}$

Substitute into the equation:

$v+X\frac{dv}{dX}=\frac{X+vX}{X-vX}=\frac{1+v}{1-v}$

Hence,

$X\frac{dv}{dX}=\frac{1+v}{1-v}-v$

Simplify the right-hand side:

$\frac{1+v-v(1-v)}{1-v}=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v}$

Thus,

$X\frac{dv}{dX}=\frac{1+v^2}{1-v}$

So,

$\frac{1-v}{1+v^2}\,dv=\frac{dX}{X}$

4. Integrate

Integrate both sides:

$\int \frac{1-v}{1+v^2}\,dv=\int \frac{dX}{X}$

Split the left side:

$\int \frac{dv}{1+v^2}-\int \frac{v\,dv}{1+v^2}=\ln|X|+C$

Now,

$\int \frac{dv}{1+v^2}=\tan^{-1}v$

and

$\int \frac{v\,dv}{1+v^2}=\frac12\ln(1+v^2)$

Therefore,

$\tan^{-1}v-\frac12\ln(1+v^2)=\ln|X|+C$

Multiply by $2$:

$2\tan^{-1}v-\ln(1+v^2)=2\ln|X|+C_1$

Since $v=\dfrac{Y}{X}$,

$2\tan^{-1}\left(\frac{Y}{X}\right)-\ln\left(1+\frac{Y^2}{X^2}\right)=2\ln|X|+C_1$

Now,

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

So the equation becomes

$2\tan^{-1}\left(\frac{Y}{X}\right)-\bigl[\ln(X^2+Y^2)-2\ln|X|\bigr]=2\ln|X|+C_1$

The $2\ln|X|$ terms cancel, giving

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

Returning to $x,y$ using $X=x-1, Y=y-1$:

$2\tan^{-1}\left(\frac{y-1}{x-1}\right)-\ln\bigl((x-1)^2+(y-1)^2\bigr)=C$

5. Use the point $(2,1)$ to find the constant

At $(x,y)=(2,1)$:

$x-1=1, \qquad y-1=0$

So,

$2\tan^{-1}(0)-\ln(1^2+0^2)=C$

$0-\ln 1=C \Rightarrow C=0$

Hence the solution curve is

$2\tan^{-1}\left(\frac{y-1}{x-1}\right)=\ln\bigl((x-1)^2+(y-1)^2\bigr)$

6. Use the second point $(k+1,2)$

At $(x,y)=(k+1,2)$:

$x-1=k, \qquad y-1=1$

Substitute into the curve equation:

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

This matches Option A.

7. Check options

• A: $2\tan^{-1}\left(\frac{1}{k}\right)=\log_e(k^2+1)$ ✔️ Correct
• B: Missing factor $2$ on LHS ✘
• C: Uses $\frac{1}{k+1}$ and wrong RHS ✘
• D: RHS is $\ln\left(\frac{k^2+1}{k^2}\right)$ which is not obtained ✘

Therefore, the correct answer is A.