Solution:
Solution
Proceeding as usual, h=1, k=1 and the equation reduces to $$\displaystyle \frac{2+v}{\left ( 1-v \right )\left ( 1+v \right )}dv= \frac{d\alpha }{\alpha }$$
Partial Fractions $$\displaystyle \left [ \frac{3}{2\left ( 1-v \right )}+\frac{1}{2\left ( 1+v \right )} \right ]dv= \frac{d\alpha }{\alpha }.$$
Integrate $$\displaystyle -3\log \left ( 1-v \right )+\log \left ( 1+v \right )= 2\log \alpha +\log k$$ or $$\displaystyle \frac{1+v}{\left ( 1-v \right )^{3}}= k\alpha ^{2}$$ or $$\displaystyle \left ( 1+\frac{\beta }{\alpha } \right )= k\alpha ^{2}\left ( 1-\frac{\beta }{\alpha } \right )^{3}$$ or $$\displaystyle \left ( \alpha +\beta \right )= k\left ( \alpha -\beta \right )^{3}$$ where $$\displaystyle \alpha = x-h= x-1, \beta = y-k= y-1$$
$$\displaystyle \left ( x+y-2 \right )= k\left ( x-y \right )^{3}.$$