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.