Question

\[xy\frac{dy}{dx} = y + 2, y\left( 2 \right) = 0\]

Answer 1

We have, 
\[xy\frac{dy}{dx} = y + 2, y\left( 2 \right) = 0\]
\[ \Rightarrow \frac{y}{y + 2}dy = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{y}{y + 2}dy = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{y + 2 - 2}{y + 2}dy = \int\frac{1}{x}dx\]
\[ \Rightarrow \int dy - 2\int\frac{1}{y + 2}dy = \log x + C\]
\[ \Rightarrow y - 2 \log \left| y + 2 \right| = \log \left| x \right| + C . . . . . (1) \]
It is given that at x = 2, y = 0 .
Substituting the values of x and y in (1), we get
\[ - 2\log 2 - \log 2 = C\]
\[ \Rightarrow - \log \left( 2^2 \times 2 \right) = C\]
\[ \Rightarrow C = - \log 8\]
Substituting the value of C in (1), we get
\[y - 2 \log \left| y + 2 \right| = \log \left| x \right| - \log 8\]
\[ \Rightarrow y - 2 \log \left| y + 2 \right| = \log \left| \frac{x}{8} \right|\]
\[\text{ Hence, }y - 2\log \left| y + 2 \right| = \log \left| \frac{x}{8} \right|\text{ is the required solution.}\]