<s>[INST] Find the closed path of given system of ordinary differential equation Consider the system of ordinary differential equations:
$$\begin{cases}\frac{dx}{dt}=4x^3y^2-x^5y^4\\
\frac{dy}{dt}=x^4y^5+2x^2y^3\end{cases}$$
Then for this system there exist 
$1).$ A closed path in $\left \{(x,y) \in \mathbb{R^2}|x^2+y^2 \leq 5  \right \}$
$2).$ A closed path in $\left \{(x,y) \in \mathbb{R^2}|5<x^2+y^2 \leq 10  \right \}$
$3). $ A closed path in $\left \{(x,y) \in \mathbb{R^2}|x^2+y^2 >10 \right \}$
$4). $ No closed Path in $\mathbb{R^2}$
solution i tried- I first find out the $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{x^2y^3+2y}{4x-x^3y^2}$$
it will become
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0\;\;\;\;\;\;\;\
....................1$$
which is of from $$f_1(xy)ydx+f_2(xy)xdy$$
 after that i find the $I.F$ of $1$ which comes out $$\frac{-1}{6xy}$$
now by multiplying this with $1$ i get $$\frac{1}{6} \left ( xy^2+\frac{2}{x} \right )  dx-\frac{1}{6} \left ( \frac{4}{y}+x^2y \right )dy=0$$
 after solving this i get answer $$\frac{x^2y^2}{12}-\frac{1}{3}\log (\frac{x}{y^2})=c$$
but there is noting related to given option ,where i am making mistake please help
Thank you
 [/INST] Starting from this line in your answer:
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0$$
$$-x^2y^2(ydx+xdy)-2ydx+4xdy=0$$
$$-x^2y^2d(xy)-2ydx+4xdy=0$$
Divide by $xy$:
$$-xyd(xy)-2\dfrac {dx}x+4\dfrac {dy}y=0$$
$$-(xy)^2-4\ln x+8\ln y=0$$
$$-(xy)^2+4\ln \dfrac {y^2}{x}=C$$
Or if you prefer:
$$(xy)^2+4\ln \dfrac x{y^2}=C_1$$
Our answers are different. There is a sign difference. This may explain AVK's comment:
There is a mistake somewhere in your solution. The picture of the phase plane of the original system does not coincide with the curves that you obtained. 

Your mistake is here: 
$$\frac{1}{6} \left ( xy^2+\frac{2}{x} \right )  dx-\frac{1}{6} \left ( \frac{4}{y}\color {red}{+x^2y} \right )dy=0$$
The term in red should be negative from the precedent line:
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0$$
You multiply by $-\dfrac 1 {6xy}$:
$$\dfrac 1 {6xy}(x^2y^2+2)ydx-\dfrac 1 {6xy}(4 \color{blue}{-x^2y^2})xdy=0$$
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