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System of differential equations question

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to solve the following system of differential equations, given some initial conditions:

    [tex]mx'' = qBy'[/tex]
    [tex]my'' = -qBx'[/tex]

    3. The attempt at a solution

    Rearranging I get [tex]x'' - \frac{qB}{m}y' = 0[/tex]
    [tex]y'' + \frac{-qB}{m}x' = 0[/tex]

    Computing the operational determinant I get [tex]cD^3 + cD^3[/tex], where C = qB/m. Continuing I apply the operators to x and y to get [tex]2cD^3x = 0[/tex] and [tex]2cD^3y = 0[/tex]. So basically I get [tex]y''' = 0[/tex] and [tex]x''' = 0[/tex]. 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. jcsd
  3. Oct 26, 2009 #2

    LCKurtz

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    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'.
     
  4. Oct 27, 2009 #3
    If I continue in that way, I find that [tex]x' = C_1 + C_2e^{-Ct}[/tex] and [tex]y' = C_3 + C_4e^{-Ct}[/tex], so x is equal to [tex]C_1t - \frac{C_2}{C}e^{-Ct} + C_3[/tex] and y is equal to [tex]C_4t - \frac{C_5}{C}e^{-Ct} + C_6[/tex]. C is equal to [tex]\frac{q^2B^2}{m^2}.[/tex]

    I now have 6 constants, but I should really only have 2...I also have 4 initial conditions from the problem, namely [tex]x_0 = r_0, y(0) = 0, x'(0) = 0, y'(0) = -\omega r_0 [/tex] 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 [tex]r_0[/tex] but I don't see it yet. Thanks so much for your help!
     
  5. Oct 27, 2009 #4

    lanedance

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

    differentiating
    [tex]x' = C_1 + C_2e^{-ct}[/tex]
    [tex]x'' = -c.C_2e^{-ct}[/tex]
    [tex]x''' = c^2 C_2e^{-ct}[/tex]

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

    follow LCKurtz suggestion let w = x' and solve
    [itex] w'' + c^2w = 0[/itex]
     
  6. Oct 27, 2009 #5
    I see it now. I made a mistake setting up my characteristic equation..:redface:
     
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