Diff Eq Solution for E&M Example: Solving y'' and z'' with Author's Method

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SUMMARY

The discussion focuses on solving a system of differential equations related to electromagnetism (E&M) using the author's method. The equations presented are y'' = ωz' and z'' = ω(E/B - y'). The author provides a solution for y(t) and z(t) involving trigonometric functions and constants. The participant confirms the approach of differentiating the first equation to eliminate z'' and seeks clarification on solving the resulting differential equation for y(t) before substituting to find z(t).

PREREQUISITES
  • Understanding of differential equations, specifically second-order equations.
  • Familiarity with trigonometric functions and their derivatives.
  • Knowledge of electromagnetism concepts, particularly the relationship between electric field (E) and magnetic field (B).
  • Proficiency in mathematical substitution techniques for solving systems of equations.
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  • Study methods for solving second-order linear differential equations.
  • Learn about the application of trigonometric identities in differential equations.
  • Explore the physical implications of the relationship between electric and magnetic fields in E&M.
  • Investigate numerical methods for solving complex differential equations when analytical solutions are challenging.
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying electromagnetism and differential equations, will benefit from this discussion.

hitmeoff
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Homework Statement


This is part of a solution to a E&M example from a book. I am in the middle of following the solution and I come across this system of diff eq's:y''=\omega z'
z''=\omega (\frac{E}{B}-y')

Homework Equations


Solution according to author is:

y(t)=C_{1}\cos (\omega t)+C_{2}\sin (\omega t)+(\frac{E}{B})t+C_{3}
z(t)=C_{2}\cos (\omega t)-C_{1}\sin (\omega t)+C_{4}

The Attempt at a Solution


So the author just says that this is easily solvable by differentiating the first and using the second to eliminate z''. Now I tried that, but correct me if I am wrong, this works out to :

z''=\frac{y'''}{\omega}
y'''+\omega^{2}y''-\omega^{2}\frac{E}{B}=0

So now do I solve this diff eq, to get y(t) and then use substitution to solve for z(t)?
 
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Sure, that looks like the right strategy.
 

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