1. Simplify and Solve the Differential Equation

The given differential equation is: $\left( {{x^2} + xy + 4x + 2y + 4} \right){{dy} \over {dx}} - {y^2} = 0, \quad x>0$ First, let's factor the coefficient of $\frac{dy}{dx}$: $x^2 + xy + 4x + 2y + 4 = (x^2 + 4x + 4) + (xy + 2y) = (x+2)^2 + y(x+2) = (x+2)(x+2+y)$ The differential equation becomes: $(x+2)(x+2+y) \frac{dy}{dx} = y^2$ This equation is not easily separable or linear in its current form. Let's try to express it as a function of $\frac{dx}{dy}$: $\frac{dx}{dy} = \frac{(x+2)(x+2+y)}{y^2}$ This is still complex. Let's introduce substitutions to simplify it. Let $X = x+2$ and $Y=y$. Then $dX = dx$ and $dY=dy$. The equation transforms to: $X(X+Y) \frac{dY}{dX} = Y^2 \implies \frac{dY}{dX} = \frac{Y^2}{X(X+Y)} = \frac{(Y/X)^2}{1+(Y/X)}$ This is a homogeneous differential equation. Let $Y=vX$, so $\frac{dY}{dX} = v + X\frac{dv}{dX}$. Substituting this into the equation: $v + X\frac{dv}{dX} = \frac{(vX)^2}{X(X+vX)} = \frac{v^2X^2}{X^2(1+v)} = \frac{v^2}{1+v}$ $X\frac{dv}{dX} = \frac{v^2}{1+v} - v = \frac{v^2 - v(1+v)}{1+v} = \frac{v^2 - v - v^2}{1+v} = \frac{-v}{1+v}$ This is a separable equation. We separate the variables: $\frac{1+v}{v} dv = -\frac{dX}{X}$ Integrating both sides: $\int \left( \frac{1}{v} + 1 \right) dv = -\int \frac{1}{X} dX$ $\ln|v| + v = -\ln|X| + C$ $\ln|vX| + v = C$ Substitute back $v = Y/X$: $\ln|Y| + \frac{Y}{X} = C$ Now substitute back $X=x+2$ and $Y=y$: $\ln|y| + \frac{y}{x+2} = C$

2. Apply the Initial Condition

The solution curve passes through the point $(1,3)$. Since $x>0$ and the given point has $y=3>0$, we can drop the absolute value sign. $\ln(y) + \frac{y}{x+2} = C$ Substitute $x=1$ and $y=3$ to find the constant $C$: $\ln(3) + \frac{3}{1+2} = C \implies \ln(3) + 1 = C$ So, the equation of the solution curve is: $\ln(y) + \frac{y}{x+2} = 1 + \ln(3)$ This can also be written as $\frac{y}{x+2} = 1 + \ln(3) - \ln(y) = 1 + \ln\left(\frac{3}{y}\right)$.

3. Analyze the Options

A: intersects $y=x+2$ exactly at one point

To find the intersection points, substitute $y=x+2$ into the solution equation: $\ln(x+2) + \frac{x+2}{x+2} = 1 + \ln(3)$ $\ln(x+2) + 1 = 1 + \ln(3)$ $\ln(x+2) = \ln(3) \implies x+2 = 3 \implies x=1$ When $x=1$, $y=x+2 = 1+2 = 3$. The intersection point is $(1,3)$. This point is in the domain $x>0$. Since the equation gives a unique value for $x$, there is exactly one intersection point. Thus, option A is correct.

B: intersects $y=x+2$ exactly at two points

As shown above, there is only one intersection point. Thus, option B is incorrect.

C: intersects $y = (x + 2)^2$

Substitute $y=(x+2)^2$ into the solution equation: $\ln((x+2)^2) + \frac{(x+2)^2}{x+2} = 1 + \ln(3)$ $2\ln(x+2) + (x+2) = 1 + \ln(3)$ Let $f(x) = 2\ln(x+2) + x+2$ for $x>0$. We need to check if $f(x) = 1 + \ln(3)$ has a solution. Let's find the derivative of $f(x)$: $f'(x) = \frac{2}{x+2} + 1$ For $x>0$, $x+2 > 2$, so $f'(x) > 0$. This means $f(x)$ is a strictly increasing function for $x>0$. The minimum value of $f(x)$ occurs as $x \to 0^+$. Let's find this minimum value: $\lim_{x\to 0^+} f(x) = 2\ln(2) + 2 = \ln(4) + 2$ The right-hand side is $1 + \ln(3)$. Let's compare the minimum value of $f(x)$ with this constant: $(\ln(4)+2) - (1+\ln(3)) = 1 + \ln(4) - \ln(3) = 1 + \ln(4/3)$ Since $4/3 > 1$, $\ln(4/3) > 0$. So, $1 + \ln(4/3) > 1 > 0$. This means $f(x)_{min} > 1+\ln(3)$. Since $f(x)$ is strictly increasing, $f(x) > 1+\ln(3)$ for all $x>0$. Therefore, there is no intersection. Option C is incorrect.

D: does NOT intersect $y = (x + 3)^2$

Substitute $y=(x+3)^2$ into the solution equation: $\ln((x+3)^2) + \frac{(x+3)^2}{x+2} = 1 + \ln(3)$ $2\ln(x+3) + \frac{(x+3)^2}{x+2} = 1 + \ln(3)$ Let $g(x) = 2\ln(x+3) + \frac{(x+3)^2}{x+2}$ for $x>0$. We need to check if $g(x) = 1 + \ln(3)$ has a solution. Let's analyze $g(x)$. We can rewrite part of it as $\frac{(x+3)^2}{x+2} = \frac{((x+2)+1)^2}{x+2} = \frac{(x+2)^2+2(x+2)+1}{x+2} = x+2+2+\frac{1}{x+2} = x+4+\frac{1}{x+2}$. So, $g(x) = 2\ln(x+3) + x+4+\frac{1}{x+2}$. Let's find the derivative of $g(x)$: $g'(x) = \frac{2}{x+3} + 1 - \frac{1}{(x+2)^2}$ For $x>0$, $x+3>3$ and $x+2>2$. Thus, $\frac{2}{x+3}>0$ and $(x+2)^2>4 \implies 0<\frac{1}{(x+2)^2}<\frac{1}{4}$. Therefore, $1 - \frac{1}{(x+2)^2} > 1 - \frac{1}{4} = \frac{3}{4} > 0$. Since both terms in $g'(x)$ are positive, $g'(x) > 0$ for $x>0$. So, $g(x)$ is a strictly increasing function. The minimum value of $g(x)$ for $x>0$ occurs as $x \to 0^+$: $\lim_{x\to 0^+} g(x) = 2\ln(3) + \frac{(0+3)^2}{0+2} = 2\ln(3) + \frac{9}{2} = 2\ln(3) + 4.5$ The right-hand side is $1 + \ln(3)$. Let's compare: $(2\ln(3)+4.5) - (1+\ln(3)) = \ln(3) + 3.5$ Since $\ln(3) \approx 1.0986 > 0$, we have $\ln(3) + 3.5 > 0$. This means $g(x)_{min} > 1+\ln(3)$. Since $g(x)$ is strictly increasing, $g(x) > 1+\ln(3)$ for all $x>0$. Therefore, there is no intersection. Option D is correct.

Conclusion

The correct options are A and D.