# Find the general solution of a coupled differential equation:

1. Sep 2, 2010

### vorcil

1. The problem statement, all variables and given/known data

I want to find the general solution of these two equations,

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

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

2. Relevant equations
These two equations are the result of quantitatively solving to find the trajectory of a charged particle in a uniform electric field, perpendicular to a magnetic one,

Suppose that the B field points in the x direction, and the E field in the z direction, and the particle is released from rest,

the lorentz force law,
$$\mathbf{F}_{mag} = Q(\mathbf{v} \times \mathbf{B})$$
3. The attempt at a solution

(i'm just going to show you how I got to the coupled differential equations

there is no force in the x direction due to the magnetic field being in that direction,

the position of the particle at t can be described by the vector, (0, y(t), z(t))

so $$\mathbf{v}=(0,\dot{y},\dot{z})$$
(the dots are the derivatives with respect to time)
calculating the cross product from the Lorentz force law,

$$\mathbf{v} \times \mathbf{B} = \left| \begin{array}{ccc} \mathbf{\hat{x}} & \mathbf{\hat{y}} & \mathbf{\hat{z}} \\ 0 & \dot{y}& \dot{z} \\ \mathbf{B}& 0& 0 \\ \end{array} \right| =B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}}$$

and applying newtons second law, F=ma=QE=Q(v\timesB)=Q(E+(v\timesB))
$$\mathbf{F} = Q(\mathbf{E}+(\mathbf{v}\times\mathbf{B})) = Q(E\mathbf{\hat{z}} + B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}} = m\mathbf{a} = m(\ddot{y}\mathbf{\hat{y}}+\ddot{z}\mathbf{\hat{z}}$$

separating the $$\hat{y} & \hat{z}$$ components

$$\mathbf{QB}\dot{z} = m\ddot{y}$$ & $$\mathbf{QE - QB}\dot{y} = m\ddot{z}$$

substituting in the cyclotron frequency formula, $$\omega = \frac{\mathbf{QB}}{m}$$

I get the two coupled differential equations that I showed earlier,

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

2: $$\ddot{z}=\omega\left(\frac{\mathbf{E}}{\mathbf{B}} - \dot{y}\right)$$

I want to figure out the general solution, but I can never get it,
I'm supposed to Differentiate the first function and then use the second to eliminate $$\ddot{z}$$

-

well if I differentiate the first equation,

I get a triple dot y, $$\dot{\ddot{y}} = w\ddot{z}$$
Is that how I differentiate the first equation?

Last edited: Sep 3, 2010
2. Sep 3, 2010

### Sourabh N

Yes. And you substitute the value of $$\ddot{z}$$ from second equation there.

3. Sep 3, 2010

### vorcil

so,

$$\dot{\ddot{y}} = \omega \omega(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y})=\omega^2 (\frac{\mathbf{E}}{\mathbf{B}}-\dot{y})$$

$$\omega^2 = (\frac{QB}{m})^2$$

$$\mathbf{QE - QB}\dot{y} = mw(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}) = QB*(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}$$

I don't understand what to do from here

4. Sep 3, 2010

### Sourabh N

Differentiate 2nd equation. Plug in y$$^{..}$$ from first equation. It looks like a sinusoidal equation, with z$$^{.}$$ as your variable. Find z$$^{.}$$ from here, and z. Plug it back in the equations and similarly find y.

The approach you are using, certainly works. Its just slightly messier.

5. Sep 3, 2010

### vorcil

ok if equation2= $$\ddot{z} = \omega (\frac{E}{B} - \dot{y})$$
then expanding

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

and differentiating

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

then plugging in $$\ddot{y}$$ from the first equation

$$\dot{\ddot{z}} = \frac{E}{B} - \omega(\omega\dot{z}) = \frac{E}{B}-\omega^2\dot{z}$$

I don't see how this looks like a sinusoidal equation?

so to get rid of those dots, I integrate?

6. Sep 3, 2010

### HallsofIvy

Well, you can't just integrate. Do you know anything at all about solving "linear differential equations with constant coefficients"?

7. Sep 3, 2010

### vorcil

Well I know how to integrate, and i've dealed with hard differential/coupled equations in physics before, but haven't started proper, differential equations in mathematics yet, I don't think?
is it just the same as integrating normal linear equations?

Last edited: Sep 3, 2010
8. Sep 3, 2010

### klondike

So it's essentially an ODE with constant coeff
$$y'''+\omega ^2y'=K$$
Perhaps the easiest way to solve it is using LapLace transform. To make life further easier, assuming you have the system completely at rest for t=0, i.e. y(0)=y'(0)=y''(0). Take LT of both sides and rearrange one would get:
$$Y(s)=\frac{K}{s^2(s^2+\omega^2)}=\frac{K}{\omega^2}(\frac{1}{s^2}-\frac{1}{s^2+\omega^2})$$

So the solution is a ramp minus a sinusoid.

9. Sep 4, 2010

### HallsofIvy

I really dislike the use of the Laplace transform for problems like this!

If you were to try a solution of the form $y= e^{rx}$, then you would have $y'= re^{rx}$, y"=$r^2e^{rx}$, $y'''= r^3e^{rx}$ and the "associated homogeneous equation" (with the "K" dropped) becomes $r^3e^{rx}+ \omega^2 r e^{rx}= (r^3+ \omega r)e^{rx}= r(r^2+ \omega)e^{rx}= 0$. Since $e^{rx}$ can't be 0, we must have $r(r^2+ \omega)= 0$ which gives r= 0 and $r= \pm i\sqrt{\omega}$. Of course, $e^{ix}= cos(x)+ i sin(x)$ so we can write $e^{it\sqrt{\omega}}$ and $it\sqrt{\omega}}$ in terms of $cos(t\sqrt{\omega})$ and $sin(t\sqrt{\omega})$. Of course, $e^{0t}= 1$: the general solution to the associated homogeneous equation is
$$C+ Dcos(t\sqrt{\omega})+ Esin(t\sqrt{\omega})$$.

To find the solution to the entire equation, try a constant solution, y= A, and add to that previous solution.

10. Sep 4, 2010

### vorcil

What should I do here Ivy? I think I'll just go down to the physics department on tuesday and ask one of the tutors to help me, I don't know how they expect me to solve this, sad thing is, I think everyone else in my class can do this really quick, but I can't, bugger it.

11. Sep 4, 2010

### klondike

Ivy's way is more "mathematician". The main idea is assuming the solution in form of complex exponential. Since God makes $$(e^{st})'=se^{st}$$, the DE will be converted to a algebra equations. I think this hinted Oliver Heaviside to invent using LT to solve this kind of DE, by letting s=a+jb.

As engineer, we were trained to use LT for linear ODE with constant coeffs which is the guts of LTI systems. We did learn Ivy's way before LT was taught. So I think you should know both.

Last edited: Sep 4, 2010