Question

Find the particular solution of the following differential equation: $xy\frac{dy}{dx}=(x+2)(y+2);y=-1$ when $x=1$.

Answer 1

Given differential equation is
$xy\frac{dy}{dx}=(x+2)(y+2)$
$\frac{y}{y+2}dy=\frac{x+2}{x}dx$
Integrating both sides,
$\int \frac{y}{y+2}dy=\int (1+\frac{2}{x}dx)$
$\Rightarrow \int (1-\frac{2}{y+2})dy=\int (1+\frac{2}{x}dx)$
$\Rightarrow y-2log|y+2|=x+2log\left|x\right|+C$
Given that y = -1 when x = 1
Therefore, $-1-2log|1|=1+2log|1|+C$
$\Rightarrow C=-2$
Therefore the required solution is
$y-2log|y+2|=x+2log|x|-2$