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प्रश्न
dy + (x + 1) (y + 1) dx = 0
उत्तर
We have,
\[dy + \left( x + 1 \right)\left( y + 1 \right) dx = 0\]
\[ \Rightarrow dy = - \left( x + 1 \right)\left( y + 1 \right) dx\]
\[ \Rightarrow \frac{1}{y + 1}dy = - \left( x + 1 \right) dx\]
Integrating both sides, we get
\[\int\frac{1}{y + 1}dy = - \int\left( x + 1 \right) dx\]
\[ \Rightarrow \log \left| y + 1 \right| = - \frac{x^2}{2} - x + C\]
\[ \Rightarrow \log \left| y + 1 \right| + \frac{x^2}{2} + x = C\]
\[\text{ Hence, }\log \left| y + 1 \right| + \frac{x^2}{2} + x =\text{ C is the required solution . }\]
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