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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.

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  • $\begingroup$ 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 $k$. $\endgroup$ Commented Apr 2, 2021 at 7:13

4 Answers 4

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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.

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  • $\begingroup$ 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
    Commented 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 $x(t)$. $\endgroup$ Commented Apr 3, 2021 at 1:17
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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.

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

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.

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

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

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