Question

\[y^2 \frac{dx}{dy} + x - \frac{1}{y} = 0\]

 

Answer 1

We have, 
\[ y^2 \frac{dx}{dy} + x - \frac{1}{y} = 0\]
\[ \Rightarrow y^2 \frac{dx}{dy} + x = \frac{1}{y} \]
\[ \Rightarrow \frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3} . . . . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dx}{dy} + Px = Q\]
where
\[P = \frac{1}{y^2}\]
\[Q = \frac{1}{y^3}\]
\[ \therefore I . F . = e^{\int P\ dy} \]
\[ = e^{\int\frac{1}{y^2}dy} \]
\[ = e^\frac{- 1}{y} \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }e^\frac{- 1}{y} ,\text{ we get }\]
\[e^\frac{- 1}{y} \left( \frac{dx}{dy} + x\frac{1}{y^2} \right) = e^\frac{- 1}{y} \frac{1}{y^3}\]
\[ \Rightarrow e^\frac{- 1}{y} \frac{dx}{dy} + x\frac{1}{y^2} e^\frac{- 1}{y} = e^\frac{- 1}{y} \frac{1}{y^3}\]
Integrating both sides with respect to y, we get
\[x\ e^\frac{- 1}{y} = \int e^\frac{- 1}{y} \frac{1}{y^3}dy + C\]
\[ \Rightarrow x\ e^\frac{- 1}{y} = I + C . . . . . . . . \left( 2 \right)\]
where
\[I = \int e^\frac{- 1}{y} \frac{1}{y^3}dy\]
\[\text{Putting }t = \frac{1}{y}, \text{ we get }\]
\[dt = - \frac{1}{y^2}dy\]

\[ = - t\int e^{- t} dt + \int\left[ \frac{d}{dt}\left( t \right)\int e^{- t} dt \right]dt\]
\[ = t e^{- t} + e^{- t} \]
\[ = \left( t + 1 \right) e^{- t} \]
\[ = \left( \frac{1}{y} + 1 \right) e^{- \frac{1}{y}} \]
\[\text{Putting the value of I in }\left( 2 \right),\text{ we get }\]
\[x\ e^\frac{- 1}{y} = \left( \frac{1}{y} + 1 \right) e^{- \frac{1}{y}} + C \]
\[ \Rightarrow x = \left( \frac{y + 1}{y} \right) + C e^\frac{1}{y} \]
\[\text{Hence, }x = \left( \frac{y + 1}{y} \right) + C e^\frac{1}{y} \text{ is the required solution.} \]