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प्रश्न
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उत्तर
\[x\frac{dy}{dx} + 1 = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 1}{x}\]
\[ \Rightarrow dy = \left( \frac{- 1}{x} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( \frac{- 1}{x} \right)dx\]
\[ \Rightarrow y = - \log\left| x \right| + C . . . . . \left( 1 \right)\]
\[\text{ It is given that }y\left( - 1 \right) = 0 . \]
\[ \therefore 0 = - \log\left| - 1 \right| + C\]
\[ \Rightarrow C = 0\]
\[\text{ Substituting the value of C in }\left( 1 \right),\text{ we get } \]
\[y = - \log\left| x \right|\]
\[\text{ Hence, }y = - \log\left| x \right|\text{ is the solution to the given differential equation .}\]
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