Question

\[xy\frac{dy}{dx} = \left( x + 2 \right)\left( y + 2 \right), y\left( 1 \right) = - 1\]

Answer 1

\[xy\frac{dy}{dx} = \left( x + 2 \right)\left( y + 2 \right), y\left( 1 \right) = - 1\]
\[ \Rightarrow \frac{y}{y + 2}dy = \frac{x + 2}{x}dx\]
\[ \Rightarrow \frac{y + 2 - 2}{y + 2}dy = \frac{x + 2}{x}dx\]
\[ \Rightarrow \left( 1 - \frac{2}{y + 2} \right)dy = \left( 1 + \frac{2}{x} \right)dx\]
Integrating both sides, we get 
\[\int\left( 1 - \frac{2}{y + 2} \right)dy = \int\left( 1 + \frac{2}{x} \right)dx\]
\[ \Rightarrow y - 2\log \left| y + 2 \right| = x + 2\log \left| x \right| + C . . . . . (1)\]
We know that at x = 1, y = - 1 . 
Substituting the values of x and y in (1), we get
\[ - 1 - 2\log \left| 1 \right| = 1 + 2\log \left| 1 \right| + C\]
\[ \Rightarrow - 1 = 1 + C\]
\[ \Rightarrow C = - 2\]
Substituting the value of C in (1), we get 
\[y - 2\log \left| y + 2 \right| = x + 2\log \left| x \right| - 2\]
\[\text{ Hence, }y - 2\log \left| y + 2 \right| = x + 2\log \left| x \right| - 2 \text{ is the required solution .} \]