Question

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

Answer 1

We have,

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

\[\Rightarrow \frac{dy}{dx} = - \frac{y^3}{xy + 1}\]

\[ \Rightarrow \frac{dx}{dy} = - \frac{xy + 1}{y^3}\]

\[ \Rightarrow \frac{dx}{dy} = - \frac{x}{y^2} - \frac{1}{y^3}\]

\[ \Rightarrow \frac{dx}{dy} + \frac{x}{y^2} = - \frac{1}{y^3}\]

\[\text{Comparing with }\frac{dx}{dy} + Px = Q,\text{ we get}\]

\[P = \frac{1}{y^2}\]

\[Q = - \frac{1}{y^3}\]

Now,

\[I . F . = e^{\int\frac{1}{y^2}dy} = e^{- \frac{1}{y}} \]

So, the solution is given by

\[x \times e^{- \frac{1}{y}} = \int - e^{- \frac{1}{y}} \frac{1}{y^3} dy + C\]

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

Now,

\[I = \int - e^{- \frac{1}{y}} \frac{1}{y^3} dy\]

\[\text{Putting }t = \frac{1}{y},\text{ we get}\]

\[dt = \frac{- 1}{y^2}dy\]

\[ = t \times \int e^{- t} dt - \int\left( \frac{d t}{d t} \times \int e^{- t} dt \right)dt\]

\[ = - t e^t + \int e^{- t} dt\]

\[ = - t e^{- t} - e^{- t} \]

\[ \therefore I = - \frac{1}{y} e^{- \frac{1}{y}} - e^{- \frac{1}{y}} = - e^{- \frac{1}{y}} \left( 1 + \frac{1}{y} \right)\]

Putting the value of `I` in (1), we get

\[x e^{- \frac{1}{y}} = - e^{- \frac{1}{y}} \left( 1 + \frac{1}{y} \right) + C\]

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