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

The differential equation of the given curve is:
$xy\frac{dy}{dx}=(x+2)(y+2)$
$⇒(\frac{y}{y+2})dy=(\frac{x+2}{x})dx$
$⇒(1-\frac{2}{y+2})dy=(1+\frac{2}{x})dx$
Integrating both sides,we get:
$∫(1-\frac{2}{y+2})dy=∫(1+\frac{2}{x})dx$
$⇒∫dy-2∫\frac{1}{y}+2dy=∫dx+2∫\frac{1}{x}dx$
$⇒y-2log(y+2)=x+2logx+C$
$⇒y-x-C=logx^2+log(y+2)^2$
$⇒y-x-C=log[x^2(y+2)^2]...(1)$
Now,the curve passes through (1,-1).
$⇒-1-1-C=log[(1)^2(-1+2)^2]$
$⇒-2-C=log1=0$
$⇒C=-2$
Substituting C=-2 in equation(1),we get:
$y-x+2=log[x^2(y+2)^2]$
This is the required solution of the given curve.