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Question
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Solution
\[\Rightarrow 2\left( y + 3 \right) = xy\frac{dy}{dx}\]
\[ \Rightarrow \frac{2}{x}dx = \frac{y}{y + 3}dy\]
\[ \Rightarrow \frac{2}{x}dx = \frac{y + 3 - 3}{y + 3}dy\]
\[ \Rightarrow \frac{2}{x}dx = \left( 1 - \frac{3}{y + 3} \right)dy\]
\[ \Rightarrow \int\frac{2}{x}dx = \int\left( 1 - \frac{3}{y + 3} \right)dy\]
\[ \Rightarrow 2\log x = y - 3\log\left| y + 3 \right| + C\]
\[ \Rightarrow \log x^2 + \log\left| \left( y + 3 \right)^3 \right| = y + C\]
\[ \Rightarrow \log\left| \left( x^2 \right) \left( y + 3 \right)^3 \right| = y + C . . . . . \left( 1 \right)\]
\[\Rightarrow \log\left| \left( 1 \right)^2 \left( - 2 + 3 \right)^3 \right| = - 2 + C\]
\[ \Rightarrow C = 2\]
Substituting the value of C in (1), we get
\[\log\left| \left( x^2 \right) \left( y + 3 \right)^3 \right| = y + 2\]
\[ \Rightarrow \left( x^2 \right) \left( y + 3 \right)^3 = e^{y + 2} \]
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