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प्रश्न
\[y^2 + \left( x + \frac{1}{y} \right)\frac{dy}{dx} = 0\]
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उत्तर
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
\[x e^{- \frac{1}{y}} = - e^{- \frac{1}{y}} \left( 1 + \frac{1}{y} \right) + C\]
\[ \Rightarrow x = - \left( 1 + \frac{1}{y} \right) + C e^\frac{1}{y}\]
