# Homework Help: System of differential equations question

1. Oct 26, 2009

### bitrex

1. The problem statement, all variables and given/known data
I have to solve the following system of differential equations, given some initial conditions:

$$mx'' = qBy'$$
$$my'' = -qBx'$$

3. The attempt at a solution

Rearranging I get $$x'' - \frac{qB}{m}y' = 0$$
$$y'' + \frac{-qB}{m}x' = 0$$

Computing the operational determinant I get $$cD^3 + cD^3$$, where C = qB/m. Continuing I apply the operators to x and y to get $$2cD^3x = 0$$ and $$2cD^3y = 0$$. So basically I get $$y''' = 0$$ and $$x''' = 0$$. I'm not sure how this helps me solve the system of differential equations, as with the characteristic equation r^3 = 0 if one root is 0 then the others could be anything...I think I've gone wrong somewhere but cannot see it. Any help would be appreciated.

2. Oct 26, 2009

### LCKurtz

Differentiating the first equation gives:

mx''' = qBy'' and from the second equation y'' = -(qB/m)x'. Substitute that in for y'':

mx''' = -(q2B2/m)x'

x''' + c2x' = 0 where c =(q2B2/m2)

Let w = x' and you have a second order constant coefficient equation in w = x'.

3. Oct 27, 2009

### bitrex

If I continue in that way, I find that $$x' = C_1 + C_2e^{-Ct}$$ and $$y' = C_3 + C_4e^{-Ct}$$, so x is equal to $$C_1t - \frac{C_2}{C}e^{-Ct} + C_3$$ and y is equal to $$C_4t - \frac{C_5}{C}e^{-Ct} + C_6$$. C is equal to $$\frac{q^2B^2}{m^2}.$$

I now have 6 constants, but I should really only have 2...I also have 4 initial conditions from the problem, namely $$x_0 = r_0, y(0) = 0, x'(0) = 0, y'(0) = -\omega r_0$$ where omega = qB/m. Do I plug the solutions for x and y back into the original equations and start equating coefficients and hope that I'll be left with only 2 constants? My goal is to show that the particle moves in a circle with radius $$r_0$$ but I don't see it yet. Thanks so much for your help!

4. Oct 27, 2009

### lanedance

how do you get $x' = C_1 + C_2e^{-ct}$ from $x''' + c^2x' = 0$?

differentiating
$$x' = C_1 + C_2e^{-ct}$$
$$x'' = -c.C_2e^{-ct}$$
$$x''' = c^2 C_2e^{-ct}$$

substituting back into the original equation gives
$$x''' + c^2x' = c^2 C_2e^{-ct} + c^2(C_1 + C_2e^{-ct}) \neq 0$$

follow LCKurtz suggestion let w = x' and solve
$w'' + c^2w = 0$

5. Oct 27, 2009

### bitrex

I see it now. I made a mistake setting up my characteristic equation..