A solution curve of the differential equation $(x^{2}+xy+4x+2y+4)\,\frac{dy}{dx}-y^{2}=0$, $x>0$, passes through the point $(1,3)$. Then the solution curve

$(x^{2}+xy+4x+2y+4)y'-y^{2}=0,\quad x>0$

$\text{passes through } (1,3)$

$\text{Q: intersections of the curve with } y=x+2,\ (x+2)^{2},\ (x+3)^{2}$

$x^{2}+xy+4x+2y+4=(x+2)^{2}+y(x+2)$

$=(x+2)(x+y+2)$

$\Rightarrow (x+2)(x+y+2)y'=y^{2}$

$s=x+2,\ u=\dfrac{s}{y}:\quad \dfrac{ds}{dy}=\dfrac{s(s+y)}{y^{2}}=\dfrac{s^{2}}{y^{2}}+\dfrac{s}{y}$

$u+y\dfrac{du}{dy}=u^{2}+u \Rightarrow \dfrac{du}{u^{2}}=\dfrac{dy}{y}$

$\Rightarrow \dfrac{y}{s}+\ln y=C \Rightarrow \dfrac{y}{x+2}+\ln y=C$

$(1,3):\ C=\dfrac{3}{1+2}+\ln 3=1+\ln 3$

$\text{Curve: } \dfrac{y}{x+2}+\ln y=1+\ln 3$

$y=x+2 \Rightarrow \dfrac{y}{x+2}=1 \Rightarrow 1+\ln y=1+\ln 3$

$\Rightarrow y=3,\ x=1: \text{ exactly one point } (1,3)$

$\Rightarrow (1) \text{ true},\ (2) \text{ false}$

$y=(x+2)^{2}:\ \dfrac{y}{x+2}+\ln y=(x+2)+2\ln(x+2)$

$x>0 \Rightarrow (x+2)+2\ln(x+2)>2+2\ln 2>1+\ln 3$

$\Rightarrow \text{no intersection} \Rightarrow (3) \text{ false}$

$y=(x+3)^{2}:\ \dfrac{y}{x+2}+\ln y=\dfrac{(x+3)^{2}}{x+2}+2\ln(x+3)$

$x>0 \Rightarrow \dfrac{(x+3)^{2}}{x+2}+2\ln(x+3)>\dfrac{9}{2}+2\ln 3>1+\ln 3$

$\Rightarrow \text{no intersection} \Rightarrow (4) \text{ true}$

$\text{Correct: } (1) \text{ and } (4)$