3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $(0,0)$.

$\endgroup$
1
$\begingroup$

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 $$x^2=-y^2+K$$ $$x^2+y^2=K$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.