- #36
- 2,168
- 193
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)Ray Vickson said: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.