Solving EM Problem with Variation of Parameters

[HPRF]

I am trying to solve the following equation using the variation of parameters method

d²x/dt²-(q²B_z²/m²)x=qE_x/m

I have put x₁=cos(t) and x₂=sin(t) into 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...

 

[LCKurtz]

Your equation is of the form x'' - a²x = b. This will have solutions of the form e^ax 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.

 

[HPRF]

This has given me the general equation

x(t)=C₁e^(qB_z/m)t + C₂e^(-qB_z/m)t

Assuming this to be correct I have the conditions x=0, v_x=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?

 

[LCKurtz]

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.

 

[HallsofIvy]

This has given me the general equation

x(t)=C₁e^(qB_z/m)t + C₂e^(-qB_z/m)t

Assuming this to be correct I have the conditions x=0, v_x=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?

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

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

That is, the solution to $d^2x/dt^2-(q^2B_z^2/m^2)x= 0$ with v(0)= 0, $v_x(0)= 0$ 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 $C_1$ and $C_2$ 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.