0
$\begingroup$

I'm trying to solve a system of coupled second order differential equations. I never did that before. I'm not sure where to begin.

The equations are:

$$\ddot{x} = -\omega\dot{y} - \frac{k}{m}\dot{x}$$ $$\ddot{y} = \omega\dot{x} - \frac{k}{m}\dot{y}$$

I'm wondering what method should I use to solve this and can I use the same method for all the coupled systems?

I know x should be: $$x(t)= -\tau V_{0x} \cos \beta e^{\frac{-t}{\tau}} \cos(\omega t + \beta) + x_0 + \tau \cos^2\beta v_{0x}$$

However, I didn't find how to process to get this solution.

I have those values at $t=0$: $$ v_x(t=0) = v_{0x} , v_y(t=0) = 0, x(t=0) = x_0, y(t=0) = y_0. $$

Finally, to lighten. $$\tau = \frac{m}{k}, \omega=\frac{qB}{m}, \tan(\beta) = \omega \tau$$

Any help would be really appreciated.

$\endgroup$
1
  • $\begingroup$ Are you sure about the derivatives in the last terms? $m\ddot x+kx=...$ is usually associated with a spring equation, without a dot on the term with factor $k$. $\endgroup$ Apr 2, 2021 at 7:13

4 Answers 4

0
$\begingroup$

The equations are:

$$x'' = -\omega y' - \frac{k}{m}x'\tag 1$$ $$y'' = \omega x' - \frac{k}{m}y'\tag 2$$

From $(1)$ $$y'=-\frac{k x'+m x''}{m \omega }\implies y''=-\frac{k x''+m x'''}{m \omega }$$

Replce both of them in $(2)$ $$\left(k^2+m^2 \omega ^2\right) x'+2 km x''+m^2 x'''=0$$ Let $p=x'$, solve for $p$ and integrate.

$\endgroup$
2
  • $\begingroup$ When you say solve for $p$, do you mean find the solution? Usually, with simple first order homogeneous differential equations I guess $p = e^{rt}$ as solution, but here I'm not sure how to get the solution for $p$. Maybe what I say doesn't make sense. $\endgroup$
    – RedDiamond
    Apr 2, 2021 at 19:33
  • $\begingroup$ @RedDiamond. Using $p=x'$ reduces the order and you face another DE (2nd order in $p$). Solve it to have $p(t)$ and since $x'(t)=p(t)$ then one more step to have $x(t)$. $\endgroup$ Apr 3, 2021 at 1:17
0
$\begingroup$

First, put $\xi = \dot x$ and $\eta = \dot y$. Then this is really a first order system: $$\frac d {dt} \begin{bmatrix} \xi\\ \eta \end{bmatrix} = \begin{bmatrix} - \omega \eta - \frac k m\xi\\ \omega \xi - \frac k m \eta \end{bmatrix} = \begin{bmatrix} -\frac k m & - \omega \\ \omega & -\frac k m\end{bmatrix}\begin{bmatrix} \xi\\ \eta \end{bmatrix}.$$ You can solve this by finding the eigenvalues and eigenvectors of the matrix. Then integrate to get $x$ and $y$. There will be a bunch of constants hanging around; you can resolve them using the initial conditions.

$\endgroup$
0
$\begingroup$

$$\ddot{x} = -\omega\dot{y} - \frac{k}{m}\dot{x}$$ $$\ddot{y} = \omega\dot{x} - \frac{k}{m}\dot{y}$$

Multiply by $\mu(t)=e^{tk/m}$: $$(\dot{x}e^{tk/m})' = -\omega\dot{y}e^{tk/m}$$ $$(\dot{y}e^{tk/m})' = \omega\dot{x}e^{tk/m}$$ This is simply: $$u' = -wv$$ $$v' = wu$$ Differentiate: $$u''=-wv'=-w^2u$$ $$u''+w^2u=0$$ This is easy to solve.

$\endgroup$
0
$\begingroup$

Define $z=x+iy$, then the system reads as $$ \ddot z=iω\dot z-\tfrac{1}{τ}\dot z. $$ This has solutions $$ z=A+B\exp((iω-\tau^{-1})t) $$ Now determine the real and imaginary components to find real expressions for $x,y$.


$z_0=A+B$, $\dot z_0=v_0=B(iω-τ^{-1})$, so that $τv_0=-B(1-i\tanβ)$, $B=-τv_0\cosβe^{iβ}$, $$ z=z_0+τv_0\cos(β)\,e^{iβ}(1-e^{-(1-i\tanβ)t/τ}) $$

$\endgroup$
1
  • $\begingroup$ Thanks for your method, it works well. I didn't know about it. $\endgroup$
    – RedDiamond
    Apr 5, 2021 at 5:14

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.