Vtyjkyuk

Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:

• $fracdxdt = (x-y)(1-x^2-y^2),$


• $fracdydt = (x+y)(1-x^2-y^2),$


• $x(0) = x_0,$


• $y(0) = y_0.$

I am asked to find $lim_ttoinftyx(t).$

Dividing the two differential equations, we have $$fracdydx = fracx+yx-y = frac1 + fracyx1-fracyx.$$
Let $v(x)=fracyx,$ so that $y = xv(x),$ and $$fracdydx = v + xfracdvdx.$$ This implies that $$xfracdvdx = frac1+v1-v - v = frac1+v^21-v.$$ Rearranging and integrating, we get $$log(x) = arctan(v)-frac12log(1+v^2)+c,$$ so $$log(x) = arctanleft(fracyxright)-frac12logleft(1+left(fracyxright)^2right)+c.$$ By writing $1+left(fracyxright)^2$ as $fracx^2+y^2x^2,$ we can simplify the above equation to $$frac12log(x^2+y^2) = arctanleft(fracyxright) + c.$$

However, I'm not sure how to proceed from this implicit equation. Any help would be much appreciated!

If you only want to compute the limit, you don't even need to solve the equation explicitly (at least for $r$).

Following @player100's suggestion, you can transform the system into
$$
begincases
dot r = r(1-r^2)\
dot theta = 1-r^2
endcases,
$$
with $r^2 = x^2+y^2$ and $theta = arctanfracyx.$

Then we can observe that our system rests at the origin as well as on the unit circle. Since $dot r > 0$ when $0 < r < 1,$ and $dot r < 0$ when $r>1$, we may conclude that any solution starting in $(x_0,y_0)$ with $x_0^2+y_0^2 > 0$ will be attracted to the unit circle, so that
$$lim_t to infty r(t) = 1.$$

Hence our limit becomes
$$ lim_tto infty x(t) = lim_tto infty r(t)costheta(t) = lim_tto infty costheta(t) = cos left(lim_tto inftytheta(t)right). $$
Now
beginalign
theta(t) - theta(t_0) &= int_t_0^t dottheta(s) , ds = int_t_0^t 1-r(s)^2 , ds,
endalign
and changing the integration variable to $u = r(s),$ so that
$$du = dot r(s) , ds = r(s)(1-r(s)^2) , ds = u(1-u^2) , ds,$$
we obtain
beginalign
int_t_0^t 1-(r(s))^2 , ds &= int_r(t_0)^r(t) frac1u , du
= log(r(t))-log(r(t_0)).
endalign
This gives
beginalign
lim_tto inftytheta(t)
&= lim_tto inftytheta(t) - theta(t_0) + theta(t_0)
= lim_tto infty log(r(t))-log(r(t_0)) + theta(t_0)\
&= log(1)-log(r(t_0)) + theta(t_0)
= arctanfracy_0x_0 - logsqrtx_0^2+y_0^2,
endalign
so that
$$lim_tto infty x(t) = cos left( arctanfracy_0x_0 - logsqrtx_0^2+y_0^2 right). $$

We first pass to polar coordinates, viz:

$x = rcos theta; ; y = r sin theta; tag 1$

$dot x = dot r cos theta - rdot theta sin theta; tag 2$

$dot y = dot r sin theta + r dot theta cos theta; tag 3$

$beginpmatrix dot x \ dot y endpmatrix = beginbmatrix cos theta & -r sin theta \ sin theta & rcos theta endbmatrix beginpmatrix dot r \ dot theta endpmatrix; tag 4$

$beginbmatrix cos theta & -r sin theta \ sin theta & rcos theta endbmatrix^-1 = dfrac1r beginbmatrix rcos theta & rsin theta \ -sin theta & cos theta endbmatrix ; tag 5$

$beginpmatrix dot r \ dot theta endpmatrix = dfrac1r beginbmatrix rcos theta & rsin theta \ -sin theta & cos theta endbmatrixbeginpmatrix dot x \ dot y endpmatrix; tag 6$

$dot x = (x - y)(1 - x^2 - y^2) = r(cos theta - sin theta)(1 - r^2); tag 7$

$dot y = (x + y)(1 - x^2 - y^2) = r(cos theta + sin theta)(1 - r^2); tag 8$

$beginpmatrix dot r \ dot theta endpmatrix = dfrac1r beginbmatrix rcos theta & rsin theta \ -sin theta & cos theta endbmatrixbeginpmatrix r(cos theta - sin theta)(1 - r^2) \ r(cos theta + sin theta)(1 - r^2)endpmatrix$
$= (1 - r^2) beginbmatrix rcos theta & rsin theta \ -sin theta & cos theta endbmatrix beginpmatrix cos theta - sin theta \ cos theta + sin theta endpmatrix = (1 - r^2) beginpmatrix r \ 1 endpmatrix; tag10$

$dot r = r(1 - r^2); tag11$

$dot theta = 1 - r^2. tag12$

Since

$x_0^2 + y_0^2 > 0, tag13$

we have

$r_0^2 = x_0^2 + y_0^2 > 0 Longrightarrow r_0 > 0; tag14$

by (11),

$0 < r < 1 Longrightarrow dot r > 0; tag15$

$1 < r Longrightarrow dot r < 0; tag16$

$r = 1 Longrightarrow dot r = 0; tag17$

it is relatively easy to see that (14)-(17) in concert imply that

$displaystyle lim_t to infty r = 1; tag18$

also, (11) and (12) together yield

$r ne 0 Longrightarrow dot(ln r) = dfracdot rr = 1 - r^2 = dot theta, tag19$

whence, integrating over $t$,

$ln r - ln r_0 = theta - theta_0, tag20$

or

$theta = ln r - ln r_0 + theta_0, tag21$

and so via (18)

$displaystyle lim_t to infty theta = lim_t to infty ln r - ln r_0 + theta_0 = theta_0 - ln r_0; tag22$

finally,

$displaystyle lim_t to infty x(t) = lim_t to infty ( r cos theta) = (lim_t to infty r)( lim_t to infty cos theta) = cos (theta_0 - ln r_0). tag23$

Let $r^2 = x^2+y^2$, $x = rcostheta$, and $y = rsintheta$.

Then, you get the following equations:
$$beginalignfracdrdt &= r(1-r^2) \
fracdthetadt &= (1-r^2)
endalign$$

The solution to the first equation is: $r^2=fracke^2t1+ke^2t$, where $k=fracr^2_01-r_0^2$.

Then, the solution to the second equation is: $theta-theta_0=log fracrr_0$,

where $r_0^2=x_0^2+y_0^2$ and $tan theta_0= fracy_0x_0$.