Checking whether solution of differential curve inersects another curve. 
The differential equation is $(x^2+xy+y^2+4x+2y+4)\dfrac {\mathrm dy}{\mathrm dx}=y^2$ passing through $(1,3) $. I have already solved it. The  solution of differential curve is:
$$\ln \left|\frac {y}{3e}\right|=-\frac {y}{x+2},x>0$$

I have to check whether the following curve intersects with these parabolas  $y=(x+2)^2$ and $y=(x+3)^2$ . Now I am not good at drawing graphs by just looking at the equation of  the curve thus I am searching for an analytical solution using calculus only. Note that I don't want the exact value, just whether the curve cuts the given two curves or not.
 A: Notice $\ln\left|\frac{y}{3e}\right| = \ln|y|- \ln(3e)$
Let $$f(x,y) = \ln|y| + \frac{y}{x+2} -\ln(3e)$$ $$x>0$$
First lets start with simply $y=x+2$. Substitute this:
$$f(x) = \ln|x+2| + \frac{x+2}{x+2} -\ln(3e)$$
If the above expression equals 0 at some point for a positive $x$ then a solution exists.
Notice the modulus disappears as we on looking for a solution when $x>0$.
$$f(x) = \ln(x+2) + 1 -\ln(3e)$$
$$f'(x) = \frac{1}{x+2}$$
Notice $$f'(x) > 0$$ when $$x>0$$
We can conclude $f(x)$ is an increasing function in the positive region.
Let us evaluate $$f(0) = \ln(2) + 1 -\ln(3e)$$
$$= \ln(2) + \ln(e) -\ln(3e)$$
$$ = \ln\frac{2}{3}$$
Recall when the argument of a logarithm (with base greater than $1$) is less than $1$ the expression is negative.
$f(x)$ is a continuous function that is strictly increasing and $f(0)$ is less than $1$ while an arbitrary positive point such as $f(7) = \ln(3)$ which is bigger than $0$. Hence an intersection exists and $1$ root exists.
Doing the same thing with $y=(x+2)^2$
Substitute this:
$$f(x) = \ln|(x+2)^2| + \frac{(x+2)^2}{x+2} -\ln(3e)$$
If the above expression equals $0$ at some point for a positive $x$ then a solution exists.
Notice the modulus disappears as we on looking for a solution when $x>0$.
$$f(x) = 2\ln(x+2) + x+2 -\ln(3e)$$
$$f'(x) = \frac{2}{x+2}+1$$
Notice $$f'(x) > 0$$ when $$x>0$$
Let us evaluate $$f(0) = 2\ln(2) + 4 -\ln(3e)$$
$$ = \ln\frac{4e^3}{3}$$
Recall when the argument of a logarithm (with base greater than $1$) is greater than $1$ the expression is positive.
$f(x)$ is a continuous function that is strictly increasing and $f(0)$ is greater than $1$. Hence no intersection exists and no root exists.
Doing the same thing with $y=(x+3)^2$
Substitute this:
$$f(x) = \ln|(x+3)^2| + \frac{(x+3)^2}{x+2} -\ln(3e)$$
If the above expression equals $0$ at some point for a positive $x$ then a solution exists.
Notice the modulus disappears as we on looking for a solution when x>0.
$$f(x) = 2\ln(x+3) + \frac{(x+3)^2}{x+2} -\ln(3e)$$
$$f'(x) = \frac{2}{x+3}+\frac{(x+3)(x+1)}{(x+2)^2}$$
Notice $$f'(x) > 0$$ when $$x>0$$
Let us evaluate $$f(0) = 2\ln(3) + \frac{9}{2} -\ln(3e)$$
$$ = \ln 3 + \frac{9}{2}$$
Recall when the argument of a logarithm (with base greater than $1$) is greater than $1$ the expression is positive.
$f(x)$ is a continuous function that is strictly increasing and $f(0)$ is greater than $1$. Hence no intersection exists and no root exists.
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