Verified answer

[tex] \large \mathbb{ \green{GIVEN : - }}[/tex]

[tex] \displaystyle\rm \: xy \frac{dy}{dx} = y + 2 [/tex]

[tex] \displaystyle\rm \: And \: \: x = 2 \: \: \: \: y = 1[/tex]

[tex] \large \mathbb{ \green{SOLUTION : - }}[/tex]

[tex] \displaystyle\rm \rightarrow \: xy \frac{dy}{dx} = y + 2 [/tex]

[tex] \displaystyle\rm \rightarrow \: \frac{y.dy}{(y + 2)} = \frac{dx}{x} [/tex]

[tex] \displaystyle\rm \rightarrow \: \frac{(y + 2 - 2).dy}{(y + 2)} = \frac{dx}{x} [/tex]

[tex] \displaystyle\rm \rightarrow \: \frac{(y + 2)dy}{(y + 2)} - \frac{2dy}{(y + 2)} = \frac{dx}{x} [/tex]

[tex] \displaystyle\rm \rightarrow \: dy - \frac{2.dy}{(y + 2)} = \frac{dx}{x} [/tex]

[tex] \rm \: On \: integrating \: both \: side[/tex]

[tex] \displaystyle\rm \rightarrow \: \int \bigg(dy - \frac{2.dy}{(y + 2)} \bigg) = \int \frac{dx}{x} [/tex]

[tex] \displaystyle\rm \rightarrow \: y - 2 \: log(y + 2) = log(x) + log(c) [/tex]

[tex] \displaystyle \pink{\rm \: at \: \: \: x = 2 \: \: \: \: y = 1}[/tex]

[tex] \displaystyle\rm \rightarrow \: 1 - 2 \: log(1 + 2) = log(2) + log(c) [/tex]

[tex] \displaystyle\rm \rightarrow \: 1 - 2 \: log(3) = log(2) + log(c) [/tex]

[tex] \displaystyle\rm \rightarrow \: 1 - \: log( {3}^{2} ) - log(2) = log(c) [/tex]

[tex] \displaystyle\rm \rightarrow \: log(e) - \: log( 9 ) - log(2) = log(c) [/tex]

[tex] \displaystyle\rm \rightarrow \: log \bigg( \frac{e}{2 \times 9} \bigg) = log(c) [/tex]

[tex] \displaystyle\rm \rightarrow \: log(c) = log(\frac{e}{18} )[/tex]

[tex] \displaystyle\rm \rightarrow \: y - 2 \: log(y + 2) = log(x) + log( \frac{e}{18} ) [/tex]

[tex] \displaystyle \boxed{\rm \rightarrow \: y = log \bigg( \frac{ex {(y + 2)}^{2}}{18} \bigg) }[/tex]