# Techniques for solving coupled second order differential equations

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.

• 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 $k$. Commented Apr 2, 2021 at 7:13

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.

• 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. Commented Apr 2, 2021 at 19:33
• @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 $x(t)$. Commented Apr 3, 2021 at 1:17

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.

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

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.

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

• Thanks for your method, it works well. I didn't know about it. Commented Apr 5, 2021 at 5:14