<s>[INST] Number of point of intersection of $y=(x+2)$ and $y=(x+2)^2$ 
The solution curve of the differential equation $\displaystyle (x^2+xy+4x+2y+4)\frac{dy}{dx}-y^2=0,x>0$
Passes through the point $(1,3).$ Then
$(1)$ number of point of intersection of  $y=x+2$
$(2)$ number of point of intersection of $y=(x+2)^2$

Try: $$\bigg[(x+2)^2+y(x+2)\bigg]\frac{dy}{dx}=y^2$$
$$y^2\frac{dx}{dy}=(x+2)^2+y(x+2)$$
Could some help me to solve it , thanks
 [/INST] $$y^2 \frac{dx}{dy} = (x+2)^2 + y(x+2)\\
\frac{1}{(x+2)^2}\frac{dx}{dy}-\frac{1}{y(x+2)} = \frac{1}{y^2}$$
Observe it is a linear differential equation by treating $\frac{-1}{x+2}$ as some variable $t$. Integration factor will be $\frac{1}{y}$ and solution is
$$\frac{-1}{(x+2)y} = \int \frac{dy}{y^3} = \frac{-1}{2y^2}-c\\
2y=(x+2)(cy^2+1)$$
I think you can proceed.
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