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
Best Answer
In this problem, it is not necessary to solve a system of differential equations. Use the Bendixson–Dulac theorem instead: $$ \frac{\partial f}{\partial x}=12x^2y^2-5x^4y^4 $$ $$ \frac{\partial g}{\partial y}=5x^4y^4+6x^2y^2 $$ The function $$ \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}=12x^2y^2+6x^2y^2=18x^2y^2, $$ has the same sign almost everywhere, therefore, the conditions of the theorem are satisfied. Hence, there are no closed orbits.