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Solving EM Problem with Variation of Parameters

  1. Nov 25, 2009 #1
    I am trying to solve the following equation using the variation of parameters method


    I have put x1=cos(t) and x2=sin(t) in to the Wronskian method. Can someone tell me if these are the correct functions to use, or should I be using exponential functions.

    Any help would be great...
  2. jcsd
  3. Nov 25, 2009 #2


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    Your equation is of the form x'' - a2x = b. This will have solutions of the form eax and e-ax for the homogeneous equation and a constant function for a particular solution of the non-homogeneous equation.

    Google "constant coefficient differential equations" in Google.
  4. Nov 25, 2009 #3
    This has given me the general equation

    x(t)=C1e(qBz/m)t + C2e(-qBz/m)t

    Assuming this to be correct I have the conditions x=0, vx=0 at t=0. These do not give me specific values for C1 or C2. It gives

    C1 = -C2 and C1 = C2 respectively.

    A specific solution is expected. Should I be looking for other conditions or am I doing something wrong?
  5. Nov 25, 2009 #4


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    You have solutions to the homogeneous equation. You need to add a particular solution to the non-homogeneous equation before you have the general solution. Then you can figure out the constants.
  6. Nov 26, 2009 #5


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    I assume you are no longer talking about the non-homogeneous equation.

    Yes, those initial conditions give [itex]C_1= -C_2[/itex] and [itex]C_1= C_2[/itex]. But why do you say they do not give a specific solution? Adding the two equations, [itex]C_2[/itex] cancels and we have [itex]2C_1= 0[/itex] so [itex]C_1= 0[/itex] and then [itex]C_2= 0[/itex]. You can't get more "specific" than that!

    That is, the solution to [itex]d^2x/dt^2-(q^2B_z^2/m^2)x= 0[/itex] with v(0)= 0, [itex]v_x(0)= 0[/itex] is the constant function x(t)= 0. It should be obvious that x(t)= 0 satisifies both differential equation and initial conditions.

    Warning: If you are still talking about the non-homogeneous equation, you cannot first find [itex]C_1[/itex] and [itex]C_2[/itex] by using the initial conditions and then add the "specific solution". The entire solution, both "homogeneous" part and "specific solution", together, must satisfy the intial conditions.

    First add the specific solution to get the entire solution, then apply the initial conditions.
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