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Equations of motion for a pendulum came out to be

$$x(t) = c_1\cos(\omega t)+c_2 \sin(\omega t)$$ $$y(t) = c_3\cos(\omega t)+c_4 \sin(\omega t)$$

How can it be shown that the trajectories of these solutions are either ellipses or lines? I think one can work directly with the two ODEs, rewrite them as a decoupled system of four linear ODEs, and then analyze their eigenvalues and sketch the phase portrait based on that.

But in this case, a simpler approach is needed. Here's what I've done so far:

$x(t) = A\cos(\omega t + \delta)$ can be rewritten as $x(\tau) = A\cos(\omega\tau)$, where $\tau := t+\frac{\delta}{\omega}$. The question is now how to rewrite $y$ in such a way is to be able to substitute $x$ into $y$ and get a suitable equation for trajectories. I took $y(\tau) = B\sin(\omega \tau)$, hence we get, after rearrangement and division, the following equation

$$ \frac{y^2}{B^2} + \frac{x^2}{A^2}=1 $$, which is clearly an ellipse. But what about lines? If $B=0$ then $y^2 + \frac{B^2}{A^2}x^2=y^2=1$, so $y=\pm 1$, which are horizontal lines. What about lines in general?

Would appreciate your feedback.

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    $\begingroup$ Well, you cannot generally take the same shifting angle for the $x$ part and the $y$ part : $$\left\{\begin{matrix}x(t)=A\cos(\omega t-\theta_x) \\ x(t)=B\cos(\omega t-\theta_y)\end{matrix}\right.$$ $\endgroup$ – Nicolas FRANCOIS Nov 15 '16 at 23:42
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You can always choose the origin of time in such a way that $c_2=0$, and reduce the scale (dividing by $c_1$), and write $$\begin{cases}x=\cos(\omega t),\\y=c\cos(\omega t)+s\sin(\omega t).\end{cases}$$

When $s=0$ you get the line segment from $(-1,-c)$ to $(1,c)$.

Otherwise, the ellipse

$$x^2+\left(\frac{y-cx}{s}\right)^2=1.$$

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  • $\begingroup$ Can you please clarify how exactly $c_2$ can become $0$ due to an origin of time? If my understanding is correct, then $c_1\cos(\omega t) + c_2 \sin(\omega t) = A\cos(\omega t - \delta) = A\cos(\delta)\cos(\omega t) + A\sin(\delta) \sin(\omega t)$ implies that we can choose the origin such where $\delta$ makes $A\sin(\delta)=0$, correct? $\endgroup$ – sequence Nov 16 '16 at 4:00
  • $\begingroup$ Make $A\cos(\omega(t-t_0)-\delta)=A\cos(\omega t)$. $\endgroup$ – Yves Daoust Nov 16 '16 at 8:00
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As \begin{align} c_4x-c_2y &= (c_1c_4-c_2c_3)\cos(ωt)\\ c_1y-c_3x &= (c_1c_4-c_2c_3)\sin(ωt) \end{align} you find that $$ (c_4x-c_2y )^2+(c_1y-c_3x )^2=(c_1c_4-c_2c_3)^2 $$ i.e., your curve is the affine image of a circle, which is in general an ellipse.

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  • $\begingroup$ You can add that when $c_1c_4-c_2c_3=0$, you get the equations of two straight lines, which are identical by this very equality. $\endgroup$ – Yves Daoust Nov 16 '16 at 0:09

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