Lorentz Force- Diff Eqn Solution

[Ray Vickson]

But one of them is $v_e$ and the other one is $v_×$

Yes, I know that. So, write $v_e = c_e e^{rt}, v_b = c_b e^{rt}$ and $v_x = c_x e^{rt}$ for some constants $c_e,c_b,c_x$.

 

[Orodruin]

I suggest differentiating the expression for $\dot{v}_e$ and inserting the expression for $\dot{v}_\times$ into the result. This should give you a second order ODE that you might recognize...

Once you have solved that the rest is just simple integration and insertion.

 

[Arman777]

Yes, I know that. So, write $v_e = c_e e^{rt}, v_b = c_b e^{rt}$ and $v_x = c_x e^{rt}$ for some constants $c_e,c_b,c_x$.

Well let me look again

 

[Arman777]

so I find $v_e=c_1sin(\sqrt{\frac {qα} {m^2}t})+c_2cos(\sqrt{\frac {qα} {m^2}t})$

for $v_x=c_1sin({\frac {q\sqrt{α}} {m}t})+c_2cos({\frac {q\sqrt{α}} {m}t})$

but for $v_x$ I guess I didnt add the non-homogeneous part which is only $q^2$.

for $v_b=\frac {qβ}{m}∫v_xdt$

 

[Ray Vickson]

so I find $v_e=c_1sin(\sqrt{\frac {qα} {m^2}t})+c_2cos(\sqrt{\frac {qα} {m^2}t})$

for $v_x=c_1sin({\frac {q\sqrt{α}} {m}t})+c_2cos({\frac {q\sqrt{α}} {m}t})$

but for $v_x$ I guess I didnt add the non-homogeneous part which is only $q^2$.

for $v_b=\frac {qβ}{m}∫v_xdt$

One way to do it is to write $v_e(t) = w_e(t) + a_e + c_e t,$, $v_b(t) = w_b(t) + a_b + c_b t,$ and $v_x(t) = w_x(t) + a_x + c_x t,$ where the a's and c's are constants. When you put those forms into the 3 DEs, you will get DEs for the w(t)s that have some constants in them. By choosing the a's and c's appropriately, you can eliminate all the constants and be left with the homogeneous system
$$
\begin{array}{rcr}
m \dot{w_e} &=& -q \alpha w_x \\
m \dot{w_b} &=& q \beta w_x \\
m \dot{w_x} &=& q w_e
\end{array}
$$
in $w_e, w_b, w_x$. The initial conditions on the $v(t)$s translate into initial conditions on the $w(t)$s, which then fixes the solution uniquely.

 

[Arman777]

One way to do it is to write $v_e(t) = w_e(t) + a_e + c_e t,$, $v_b(t) = w_b(t) + a_b + c_b t,$ and $v_x(t) = w_x(t) + a_x + c_x t,$ where the a's and c's are constants. When you put those forms into the 3 DEs, you will get DEs for the w(t)s that have some constants in them. By choosing the a's and c's appropriately, you can eliminate all the constants and be left with the homogeneous system
$$
\begin{array}{rcr}
m \dot{w_e} &=& -q \alpha w_x \\
m \dot{w_b} &=& q \beta w_x \\
m \dot{w_x} &=& q w_e
\end{array}
$$
in $w_e, w_b, w_x$. The initial conditions on the $v(t)$s translate into initial conditions on the $w(t)$s, which then fixes the solution uniquely.

I see your point I guess but it comes the same way as I did right i mean. Your equations solutions and my solutions has to be same(if they are correct)