Question

For the differential equation $xy\frac{dy}{dx}=(x+2)(y+2),$ find the equation of the curve passing through the point (1, -1).

Answer 1

The given curve is $xy\frac{dy}{dx}=(x+2)(y+2)$ ....(i)
On separating the variables, we get $\frac{y}{y+2}dy=\frac{x+2}{x}dx$
On integrating both sides, we get
$\int \frac{y}{y+2}dy=\int \frac{x+2}{x}dx\Rightarrow \int \left(\frac{y+2-2}{y+2}\right)dy=\int \frac{x+2}{x}dx\Rightarrow \int \frac{y+2}{y+2}dy-\int \frac{2}{y+2}dy=\int \frac{x}{x}dx+\int \frac{2}{x}dx\Rightarrow \int dy-2\int \frac{1}{y+2}dy=\int dx+2\int \frac{1}{x}dx\Rightarrow y-2log|y+2|=x+2log|x|+C....\left(ii\right)$
Now, the curve passes through the point (1,-1).
so, put x=1 and y=1
-1-2log|-1+2|=1+2log1 +C
$\Rightarrow -1-2log1=1+2log1+C\Rightarrow C=-2[\because log1=0]$
Hence, from Eq.(ii) we get
$y-2log|y+2|=x+2log|x|-2\Rightarrow y-x+2=2log|x(y+2)|\Rightarrow y-x+2=log\left[x^2\right(y+2{)}^2]$
which is the required equation of the curve.