# Finding the integral curves of a vector field $x\partial y- y\partial x$

well, i know that the curves is (x,y) where

$\displaystyle\frac{dx}{dt}=-y$

and

$\displaystyle\frac{dy}{dt}=x$

i know that this is a PDE system, my professor said that if use second derivatires it may be turned into the ODE problem

$\displaystyle\frac{d^{2}x}{dt^2}=-x$

and its solution is a linear combination of $sin$ and $cos$, but this talk is too abstract for me, i would like if someone give me a exact solution of this problem. Is there another way, easier to solve this?

Multiply $$\displaystyle\frac{dx}{dt}=-y$$ by $x$ to get $$\displaystyle x\frac{dx}{dt}=-xy$$

Multiply $$\displaystyle\frac{dy}{dt}=x$$ by $y$ to get $$\displaystyle y\frac{dy}{dt}=xy$$

Add them together to get $$x \frac{dx}{dt} +y \frac{dy}{dt} =0$$

Integrate to get $$x^2 + y^2 =C$$

That is the integral curve for your system.

Perhaps it might be more clear if you write $y = y(t)$ and $x = x(t)$. Then you get $x'(t) = -y(t)$ and $y'(t) = x(t)$.

As you wrote, substituting one of the equalities into the other gets us $x''(t) = - y'(t) = -x(t)$.

Similarly, you have $y''(t) = x'(t) = -y(t)$. Thus, you get the two equations: $$x''(t) + x(t) = 0$$ $$y''(t) + y(t) = 0$$ The characteristic equation associated to these is $r^2 + 1 = 0$. Let's just look at the differential equation associated with $x(t)$. This has complex roots, so the solution should be of the form $x(t) = c_1 \cos(t) + c_2 \sin(t)$ for arbitrary constants $c_1$ and $c_2$.

So, $y(t) = c_1 \sin(t) - c_2 \cos(t)$.

Writing the curve as $\gamma(t)$, we have $\gamma(t) = (c_1 \cos(t) + c_2 \sin(t), c_1 \sin t - c_2 \cos t)$ which gives us circles except at the point $(0,0)$.

you can multiply the first equation by the second one $$\frac {dx}{dt}=-y$$ $$\frac {dx}{dt}\frac {dt}{dy}=-\frac yx$$ $$\frac {dx}{dy}=-\frac yx$$ $$x{dx}=-y{dy}$$ Integrate $$x^2=-y^2+K$$ $$x^2+y^2=K$$