The given equation is:

\({{\rm{y}}^2}{\rm{dx}} + \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}} = 0\)

\(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = - \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = \left( {\frac{1}{{\rm{y}}} - {\rm{x}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{y}}^2}\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{\rm{y}}} - {\rm{x}}\)

\(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^3}}} - \frac{{\rm{x}}}{{{{\rm{y}}^2}}}\)

\(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} + \frac{{\rm{x}}}{{{{\rm{y}}^2}}} = \frac{1}{{{{\rm{y}}^3}}}\)

The general form of first order of differentiation is:

\(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\)

On comparing,

\(\Rightarrow {\rm{P}} = \frac{1}{{{{\rm{y}}^2}}} = {{\rm{y}}^{ - 2}}\)

\(\Rightarrow {\rm{Q}} = \frac{1}{{{{\rm{y}}^3}}}\)

Now, the integration factor is given by the formula:

I = e∫Pdy

Now,

⇒ ∫P dy = ∫(y-2)dy

\(\Rightarrow \smallint {\rm{Pdy}} = - {{\rm{y}}^{ - 1}} = - \frac{1}{{\rm{y}}}\)

Now, the integration factor for the given equation is:

\(\Rightarrow {\rm{I}} = {{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}\)

Now, the general solution is given by the formula:

Ix = ∫IQdy

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^3}}}} \right){\rm{dy}}\)

On putting \({\rm{t}} = - \frac{1}{{\rm{y}}} = - {{\rm{y}}^{ - 1}}\)

\(\Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^2}}}\)

\(\Rightarrow {\rm{dt}} = \frac{{{\rm{dy}}}}{{{{\rm{y}}^2}}}\)

Now,

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^2}}}.\frac{1}{{\rm{y}}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \smallint \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}\)

This is integration of uv which is given by the formula:

⇒ ∫u dv = uv - ∫v du

Now,

⇒ u = t and dv = et

\(\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dt}}}} = 1{\rm{\;and\;v}} = {{\rm{e}}^{\rm{t}}}\)

\(\Rightarrow {\rm{du}} = {\rm{dt\;and\;v}} = {{\rm{e}}^{\rm{t}}}\)

On substituting,

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - \smallint \left( {{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}} \right)\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - {{\rm{e}}^{\rm{t}}}} \right)\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}} - {\rm{t}}.{{\rm{e}}^{\rm{t}}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}}\left( {1 - {\rm{t}}} \right) + {\rm{C}}\)

On putting \({\rm{t}} = - \frac{1}{{\rm{y}}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}\left( {1 + \frac{1}{{\rm{y}}}} \right) + {\rm{C}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}\left( {\left( {1 + \frac{1}{{\rm{y}}}} \right) + \frac{{\rm{C}}}{{{{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}}}} \right)\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \frac{{\rm{C}}}{{{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}}}\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + {\rm{C}}{{\rm{e}}^{\frac{1}{{\rm{y}}}}}\)

Given, y = 1 and x = 1,

\(\Rightarrow 1 = 1 + \frac{1}{1} + {\rm{C}}{{\rm{e}}^{\frac{1}{1}}}\)

⇒ 1 = 2 + Ce1

⇒ Ce = 1 – 2

⇒ Ce = -1

\(\therefore {\rm{C}} = - \frac{1}{{\rm{e}}}\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \left( { - \frac{1}{{\rm{e}}}} \right){{\rm{e}}^{\frac{1}{{\rm{y}}}}}\)

The value of y = 2, the value of ‘x’ is:

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{2} + \left( { - \frac{1}{{\rm{e}}}} \right){{\rm{e}}^{\frac{1}{2}}}\)

\(\Rightarrow {\rm{x}} = \frac{{2 + 1}}{2} - \left( {\frac{1}{{\rm{e}}} \times {{\rm{e}}^{\frac{1}{2}}}} \right)\)

\(\Rightarrow {\rm{x}} = \frac{3}{2} - \left( {{{\rm{e}}^{ - 1}}.{{\rm{e}}^{\frac{1}{2}}}} \right) = \frac{3}{2} - \left( {{{\rm{e}}^{\left( { - 1 + \frac{1}{2}} \right)}}} \right) = \frac{3}{2} - {{\rm{e}}^{ - \frac{1}{2}}} = \frac{3}{2} - \frac{1}{{{{\rm{e}}^{1/2}}}}\)