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प्रश्न
y (1 + ex) dy = (y + 1) ex dx
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उत्तर
We have,
\[y\left( 1 + e^x \right) dy = \left( y + 1 \right) e^x dx\]
\[ \Rightarrow \frac{y}{y + 1}dy = \frac{e^x}{1 + e^x}dx\]
Integrating both sides, we get
\[\int\frac{y}{y + 1}dy = \int\frac{e^x}{1 + e^x}dx\]
\[\text{ Substituting }1 + e^x = t, \text{ we get }\]
\[ e^x dx = dt\]
\[ \therefore \int\frac{y}{y + 1}dy = \int\frac{1}{t}dt\]
\[ \Rightarrow \int\frac{y + 1 - 1}{y + 1}dy = \int\frac{1}{t}dt\]
\[ \Rightarrow \int dy - \int\frac{1}{y + 1}dy = \int\frac{1}{t}dt\]
\[ \Rightarrow y - \log \left| y + 1 \right| = \log \left| t \right| + C\]
\[ \Rightarrow y - \log \left| y + 1 \right| = \log \left| 1 + e^x \right| + C\]
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