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I have encountered this set of equation in a book explaining the cycloid motion of particle in orthogonal magnetic and electric field. In usual coordinate system.

$y''(t)=pz'(t)$

$z''(t)=p(q-y'(t))$

where p and q are constants and $y(t)$, $z(t)$ are position of particle in space at time t.Their general solution they written is

$y(t)=C_1\cos(pt)+C_2\sin(pt)+qt+C_3$

$z(t)=C_2\cos(pt)-C_1\sin(pt)+C_4$

which i am unable to solve. Please help.

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    $\begingroup$ You do have the general solution. What more do you need ? $\endgroup$ – Yves Daoust Jun 16 '15 at 9:56
  • $\begingroup$ i want to know how to solve these types? $\endgroup$ – bra Jun 16 '15 at 9:58
  • $\begingroup$ Do you mean how you can get the general solution? $\endgroup$ – Ruvi Lecamwasam Jun 16 '15 at 9:59
  • $\begingroup$ of course andrew $\endgroup$ – bra Jun 16 '15 at 10:00
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    $\begingroup$ Are you able to solve a single ordinary linear differential equation ? $\endgroup$ – Yves Daoust Jun 16 '15 at 10:01
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from equation 1 $$z'=\frac{y''}{p}$$ $$z''=\frac{y'''}{p}$$ substitute in the second equation $$\frac{y'''}{p}=pq-py'$$ $$y'''+p^2y'=p^2q$$ integrate both sides $$y''+p^2y=p^2qt+K_3$$ the particular solution is $$y_p=C_1\cos pt+C_2\sin pt$$ to find the complementary solution we will assume $$y_c=At+B$$ substitute to get $$A=q$$ $$B=\frac{K_3}{p^2}=C_3$$ So $$y=y_p+y_c$$ $$y=C_1\cos pt+C_2\sin pt+C_3+q*t$$ then you can use this solution to find the $z$

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  • $\begingroup$ @bra you are welcome $\endgroup$ – E.H.E Jun 16 '15 at 10:10
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The solution of linear systems is similar to that of single equations.

First let us lower the degree of the equations by setting $u:=y',v:=z'$ and rewrite the system in a canonical form:

$$u'(t)-pv(t)=0\\v'(t)+pu(t)=pq.$$

Find the general solution of the homogenous system

$$u'(t)-pv(t)=0\\v'(t)+pu(t)=0.$$

You can do that by trying an exponential solution like $u=u_0e^{\lambda t},v=v_0e^{\lambda t}$. You get:

$$\lambda u_0e^{\lambda t}-pv_0e^{\lambda t}=0\\\lambda v_0e^{\lambda t}+pu_0e^{\lambda t}=0,$$ i.e. after simplification $$\lambda u_0-pv_0=0\\\lambda v_0+pu_0=0.$$ You should recognize an Eigenproblem for a $2\times2$ matrix (two elements are missing). The Eigenvalues are $\pm ip$, and corresponding Eigenvectors are $(1,i), (1,-i)$.

Combining, the general solution can be written $$u(t)=C_0\cos(pt)+C_1\sin(pt)\\v(t)=C_1\cos(pt)-C_0\sin(pt).$$

Now looking at the non-homogenous system, we should find a constant particular solution:

$$u(t)=u_0\\v(t)=v_0.$$ Plugging in the system we get

$$-pv_0=0\\pu_0=pq,$$ and by identification, $$u_0=q,v_0=0.$$ Now, to get $y$ and $z$, you combine and integrate once on $t$.

This is a quick summary of a general approach to solve such systems, which can be based on matrix algebra techniques.

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From the first equasion you have $y'(t) = pz(t)+D_1. (D_1$ is constant). Substitute this to the second and have $z''(t)+p^2(z(t)+\frac{D_1-q}{p})=0$. This is second-order linear ordinary differential equation with general solution you written. Then solve for y(t). Take care of constants to arrive q be coefficient of t in y(t).

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