Official solution
\[{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0\]
\[ \Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}\]
Integrating factor (I.F) = \[{e^{ - {1 \over y}}}\]
Now x.\[{e^{ - {1 \over y}}}\] = \[\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} \] by putting \[{ - {1 \over y}}\] = t
x.et = \[\int {{e^t}( - t)dt} \]
\[ \Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c\]
\[ \Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c\]
\[ \Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}\]
It passes through (1, 1)
\[ \therefore \] c = \[-{1 \over e}\]
Equation of the curve is
\[x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}\]
It passes through (k, 2)
\[ \therefore \] k = \[1 + {1 \over 2} - {e^{{1 \over 2} - 1}}\]
= \[3 \over 2\] - \[1 \over{\sqrt e}\]