Question

For the differential equation xy \[\frac{dy}{dx}\] = (x + 2) (y + 2). Find the solution curve passing through the point (1, −1).

Answer 1

We have,

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

\[ \Rightarrow \frac{y}{y + 2}dy = \frac{\left( x + 2 \right)}{x}dx\]

Integrating both sides, we get

\[\int\frac{y}{y + 2}dy = \int\frac{\left( x + 2 \right)}{x}dx\]

\[ \Rightarrow \int d y - 2\int\frac{1}{y + 2}dy = \int dx + 2\int\frac{1}{x}dx\]

\[ \Rightarrow y - 2 \log \left| y + 2 \right| = x + 2 \log \left| x \right| + C . . . . . (1)\]

This equation represents the family of solution curves of the given differential equation.

We have to find a particular member of the family, which passes through the point (1, - 1).

Substituting x = 1 and y = - 1 in (1), we get

\[ - 1 - 2 \log \left| 1 \right| = 1 + 2 \log \left| 1 \right| + C\]

\[ \Rightarrow C = - 2\]

Putting `C =-2` in (1), we get

\[ y - 2 \log \left| y + 2 \right| = x + 2 \log \left| x \right| - 2 \]

\[ \Rightarrow y - x + 2 = \log \left\{ x^2 \left( y + 2 \right)^2 \right\} \]

\[\text{Hence, }y - x + 2 = \log \left\{ x^2 \left( y + 2 \right)^2 \right\} \text{ is the equation of the required curve.}\]