Update 3_drude_model_solutions.md - typo fix

src/3_drude_model_solutions.md

@@ -47,19 +47,17 @@ $$
 \frac{dv_y}{dt} = \frac{ev_xB_z}{m}
 $$
 
-
-
 2.
 
-Compute $v_x(t)$ and $v_y(t)$ by solving the differential equations in 1. 
+We can compute $v_x(t)$ and $v_y(t)$ by solving the differential equations in 1.
 
 From 
 $$
 v_x'' = -\frac{e^2B_z^2}{m^2}v_x
 $$
-and the initial conditions we find $v_x(t) = v_0 \cos(\omega_c t)$ with $\omega_c=eB_z/m$. From this we can derive $v_y(t)=-v_0\sin(\omega_c t)$. 
+and the initial conditions, we find $v_x(t) = v_0 \cos(\omega_c t)$ with $\omega_c=eB_z/m$. From this we can derive $v_y(t)=v_0\sin(\omega_c t)$. 
 
-We can now calculate the positions using $x(t)=x_0 + \int_0^t v_x(t')dt'$, and similarly for $y$, to find a parametric expression for the position of the particle 
+We can now calculate the particle position using $x(t)=x_0 + \int_0^t v_x(t')dt'$ (and similar for $y(t)$). We find a parametric expression for the position of the particle 
 $$
 (x(t) - x_0)^2 + (y(t) - y_0)^2 = \frac{v_0^2}{\omega_c^2}
 $$
@@ -71,13 +69,13 @@ Due to the applied electric field $\bf E$ in the $x$-direction, the equations of
 $$
 m v_x' = -e(E_x + v_yB_z)
 $$
-This leads to the same 2nd order D.E. for v_x as above. However, for $v_y$ we get
+This leads to the same 2nd-order D.E. for v_x as above. However, for $v_y$ we get
 $$
-v_y'' = -\omega_c^2(\frac{E_x}{B_z}+v_y)
+v_y'' = -\omega_c^2(\frac{E_x}{B_z}+v_y),
 $$ 
-leading to 
+which has as solution 
 $$
-v_y(t) = -v_0\sin(\omega_t)-\frac{E_x}{B_z}
+v_y(t) = v_0\sin(\omega_t)-\frac{E_x}{B_z}.
 $$
 By integrating the expressions for the velocity we find:
 $$