Update 3_drude_model_solutions.md - typo fix

src/3_drude_model_solutions.md

@@ -75,13 +75,13 @@ v_y'' = -\omega_c^2(v_d+v_y),
 $$ 
 where we defined $v_d=\frac{E_x}{B_z}$. The general solutions are
 $$
-v_y(t) = c_1\sin(\omega_c t)+ c_2\cos(\omega_c t) -$v_d$. \\
-v_x(t) = c_3\sin(\omega_c t)+ c_4\cos(\omega_c t)
+v_y(t) = c_1\sin(\omega_c t)+ c_2\cos(\omega_c t) -v_d \\
+v_x(t) = c_3\sin(\omega_c t)+ c_4\cos(\omega_c t).
 $$
 Using the initial conditions $v_x(0)=v_0$ and $v_y(0)=0$ and the 1st order D.E. above, we can show
 $$
-v_y(t) = v_0\sin(\omega_c t)+ v_d\cos(\omega_c t) -v_d. \\
-v_x(t) = v_d\sin(\omega_c t)+ v_0\cos(\omega_c t) \\
+v_y(t) = v_0\sin(\omega_c t)+ v_d\cos(\omega_c t) -v_d \\
+v_x(t) = v_d\sin(\omega_c t)+ v_0\cos(\omega_c t).
 $$
 
 By integrating the expressions for the velocity we find: