$\ddot{y}=\omega \dot{z}$

$\ddot{z}=\omega (\frac{E}{B}-\dot{y})$

These are the coupled differential equation i came across . They have already been asked here

How to Solve the Coupled Differential Equations?

but i am not satisfied with answer because i can not proceed further. Please give me its complete solution.


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  • $\begingroup$ This question has its source in Griffiths, Introduction to Electrodynamics, Second Edition, page 200. $\endgroup$ Apr 28, 2021 at 17:34

2 Answers 2


Just as, in solving a system of linear equations, one reduces to a single equation in one unknown, so in a system of linear differential equations, one reduces to a single equation, of higher order.

The first of these equations says that $y''= \omega z'$. Differentiating again, $y'''= \omega z''$ so that $z''= \frac{y'''}{\omega}$. Replace z'' in the second equation by that: $\frac{y'''}{\omega}= \omega(\frac{B}{E}- y')$ so that $y'''= \omega^2(\frac{E}{B}- y')= \frac{\omega^2 E}{B}- \omega^2 y'$

That is the "linear, non-homogeneous, differential equation with constant coefficients" $y'''+ \omega^2 y'= \frac{\omega^2 E}{B}$.

The "associated homogeneous equation" is $y'''+ \omega^2 y'= 0$ which has characteristic equation $r^3+ \omega^2 r= r(r^2+ \omega^2)= r(r+ i\omega)(r- i\omega)= 0$. That has roots 0, $i\omega$, and $-i\omega$ so then general solution to the associated homogeneous equation is $y(t)= A+ Bcos(\omega t)+ C sin(\omega t)$.

The "non-homogenous" part of the equation is then constant, $\omega\frac{E}{B}$. Normally, because that is a constant, we would "try" a constant as a solution but, because the constant, A, is a solution to the associated homogeneous equation, new try $y= Dt$ for D a constant. Then $y'= D$, $y''= 0$, and $y'''= 0$ so the equation becomes $\omega^2 D= \frac{\omega^2 E}{B}$ so $D= \frac{E}{B}$ and the general solution to the entire equation is $y(t)= A+ Bcos(\omega t)+ C sin(\omega t)+ \frac{E}{B}$.

  • $\begingroup$ There is one difference between your solution, and the solution given by Griffithsion given by Griffiths. Griffiths. Griffiths has $y(t) = C_1\,\cos(\omega\,t)+C_2\,\sin(\omega\,t) + \frac{E}{B}\,t + C_3.$. One of you has an error. Either you are missing a $t$, or Griffiths has one too many $\endgroup$ Apr 28, 2021 at 17:37

Continuing with the solution of How to Solve the Coupled Differential Equations?

The general solution of : $$\ddot q + \omega^2q=\omega^2\frac{E}{B}$$ is : $$q(t)=C_1\cos(\omega t)+C_2\sin(\omega t) +\frac{E}{B}$$

then $$y(t)=\frac{C_1}{\omega}\sin(\omega t)-\frac{C_2}{\omega}\cos(\omega t)+\frac{E}{B}t+C_3$$ Now I guess that you can find $z$

  • $\begingroup$ You have been too fast for the old man ! I was typing the same when your answer came. Cheers. $\endgroup$ Feb 13, 2016 at 7:52
  • $\begingroup$ @ClaudeLeibovici often happens to me too $\endgroup$
    – stity
    Feb 13, 2016 at 7:54
  • $\begingroup$ In general solution we have $iw$ so we get sin and cos terms . Please tell me how we get $\frac{E}{B}$ term $\endgroup$
    – hood
    Feb 13, 2016 at 12:20
  • $\begingroup$ @ Claude Leibovici Dynamic problem in Reals :) $\endgroup$
    – Narasimham
    Feb 13, 2016 at 13:39

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