Coordinates of a particle (coupled differential equations)

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SUMMARY

The discussion focuses on solving coupled differential equations for the coordinates of a particle in the xy plane, specifically \(\dot{x} + \omega y = 0\) and \(\dot{y} - \omega x = 0\). Participants derive the equation \(\dot{z} - i\omega z = 0\) by substituting \(\dot{z} = \dot{x} + i\dot{y}\) and expressing \(\dot{x}\) and \(\dot{y}\) in terms of \(x\) and \(y\). The solution leads to \(z = z_0 e^{\omega t}\), where \(z_0\) is a complex integration constant that incorporates a phase angle \(\phi\), resulting in the final form \(z = r_0 e^{i(\omega t + \phi)}\).

PREREQUISITES
  • Understanding of complex variables, specifically \(z = x + iy\)
  • Familiarity with differential equations and their solutions
  • Knowledge of Euler's formula and complex exponentials
  • Basic concepts of phase angles in complex numbers
NEXT STEPS
  • Study the derivation of complex variables in differential equations
  • Learn about Euler's formula and its applications in solving differential equations
  • Explore the role of phase angles in complex number solutions
  • Investigate the properties of exponential functions in the context of complex analysis
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Students and educators in mathematics, particularly those studying differential equations and complex analysis, as well as anyone interested in the application of these concepts in physics and engineering.

Linday12
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Homework Statement


I'm given two equations for coordinates of a certain particle in the xy plane, \dot{x}+ωy=0
and \dot{y}-ωx=0.
Then using the complex variable z=x+iy, find the differential for z, and solve it. Hence give x and y as functions of time.


Homework Equations




The Attempt at a Solution


I'm not sure how this is gotten:
\dot{z}-iωz=0

Any help would be highly appreciated. Thank you!
 
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Hi Linday12! :smile:

Can you write \dot z in terms of x and y?
And then substitute the equations you have?

Can you also write iωz in terms of x and y?
 
Hi. I'm not really sure what to do. As far as I know, taking \dot{z}=\dot{x}+i\dot{y}, and \dot{y}=ωx and \dot{x}=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.
 
Linday12 said:
Hi. I'm not really sure what to do. As far as I know, taking \dot{z}=\dot{x}+i\dot{y}, and \dot{y}=ωx and \dot{x}=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.

So substitute your 2nd and 3rd equation in the first?
 
Yes. I then get \dot{z}=-ωy+iωx=ω(-y+ix). So I can sort of see the relation here now, except my variables seem to be the wrong way, and I have no idea where the \dot{z}-iωz=o comes from still, because i would be gone if the variables were the proper way and I subbed in z for the x-iy.
 
What is iz in terms of x and y?
 
Wow, interesting. So \dot{z}=iωz → \dot{z}-iωz=0 because iz=ix-y. Thank you! Sorry, one more question. Now that I have that, when solving it I get z=z_{0}e^{ωt}, but the answer has an additional phase angle \phi in it, so I was just wondering if there was a quick explanation for that.

It's been a while since I did any differentials (I haven't had a class in them yet), I think perhaps I'm used to the general formula that includes the phase angle in the constant z_0, but if you were to solve it you would get ln(z)=iωt+c, which is where the phase angle came from, but then where does the z_0 come from? Because the answer is z=z_0e^{ωt+\phi} Sorry, I hope that makes sense.
 
Your z0 is the integration constant that is itself a complex number.
Write z_0=r_0e^{i\phi} and you get:
z=z_0e^{i\omega t}=r_0e^{i\phi}e^{i\omega t}=r_0e^{i(\omega t + \phi)}

Btw, note the extra i and the use of the non-negative real number r0 instead of z0.
 

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