A solution of the differential equation

$\left.(x^2+xy+4x+2y+4\right)\frac{dy}{dx}-y^2=0,x>0$  passes through the point $(1 ,3)$. The solution curve 

The given differential equation is  

$\begin{array}{l}\left(x^2+xy+4x+2y+4\right)\frac{dy}{dx}-y^2=0,x>0\\ \Rightarrow \left\{\left(x^2+4x+4\right)+(xy+2y)\right\}\frac{dy}{dx}=y^2\\ \Rightarrow \left\{(x+2{)}^2+y(x+2)\right\}\frac{dy}{dx}=y^2\\ \Rightarrow (x+2)(x+y+2)\frac{dy}{dx}=y^2\end{array}$

$\Rightarrow \frac{dy}{dx}=\frac{y^2}{(x+2)(x+2+y)}$                  …(i)

Let $x+2=X$ and $y=Y\text{. }$Then , $\frac{dy}{dx}=\frac{dy}{dY}\frac{dY}{dX}\frac{dX}{dx}=\frac{dY}{dX}$

Substituting these values in (i), we get 

$\begin{array}{l}\frac{dY}{dX}=\frac{Y^2}{X(X+Y)}\\ \Rightarrow \left(X^2+XY\right)dY=Y^{2}dX\\ \Rightarrow X^{2}dY=Y^{2}dX-XYdY\\ \Rightarrow -\frac{1}{Y}dY=\frac{XdY-YdX}{X^2}\\ \Rightarrow \frac{1}{Y}dY=d\left(\frac{Y}{X}\right)\end{array}$ 

On integrating, we get  

$-\log \left|Y\right|=\frac{Y}{X}+C\Rightarrow -\log \left|y\right|=\frac{y}{x+2}+C$              …(ii)

The curve given in (ii) passes through the point (1, 3). 

$-\log 3=1+C\Rightarrow C=-\log 3-1$

Substituting the value of C in (ii), we obtain 

$-\log \left|y\right|=\frac{y}{x+2}-\log 3-1$                     …(iii)

In order to find the points of intersection (if any) of (iii) and  $y=x+2,$ we solve the two equations simultaneously.

 Putting $y=x+2$in (iii), we obtain 

$-\log (x+2)=-\log 3\Rightarrow x+2=\pm 3\Rightarrow x=1.[\because x>0]$

Putting $x = 1$ in $y=x+2$, we obtain $y = 3.$

So, the solution curve (iii) intersects $y=x+2$ exactly at one point $(1,3)$. 

So, option (a) is correct but option (b) is incorrect. Putting  $y = (x + 2{)}^2$ in (iii), we obtain 

$-2\log |x+2|=(x+2)-\log 3-1$

$\begin{array}{l}\Rightarrow 2\log (x+2)=-x-1+\log 3\\ \Rightarrow x+1+2\log (x+2)=\log 3\end{array}$

We find that $f\left(x\right) =x+1+2 \log (x + 2)$ is strictly increasing for $x > 0$ and its minimum value is $f\left(0\right) = 1 + 2 \log 2 > \log 3$ . 

So, the above equation has no solution. 

So, $y = {(x + 2)}^2$ does not intersect the solution curve. 

To find the point of intersection of the solution curve (iii) and

$y=(x+3{)}^2,$we put $y=(x+3{)}^2$ in (iii) to get 

$-\log (x+3{)}^2=\frac{(x+3{)}^2}{x+2}-\log 3-1$

$\Rightarrow \log \frac{(x+3{)}^2}{3}+\frac{(x+3{)}^2}{x+2}-1=0$

For $x>0,\log \frac{(x+3{)}^2}{3}+\frac{(x+3{)}^2}{x+2}-1>0$ So, the above equation has no solution.

 Hence,$y = {(x + 3)}^2$ does not intersect the solution curve.