Finding components of vector in the form of ODE problem

Question

The vector $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ satisfies the vector equation

$m\frac{d^2r}{dt^2}=e\mathbf{E}+\frac{e}{c}(\frac{d\mathbf{r}}{dt} \wedge \mathbf{H})$

where $\mathbf{E}=(0,E,0), \mathbf{H}=(0,0,H)$, and

$e,m,c,E,H $ are constants. Wr/ite the equation in component form and show by solving the equation that

$x=\frac{cEt}{H}-\frac{mc^2E}{eH^2}\sin{(\frac{eH}{mc}t)}$,

$y=\frac{mc^2E}{eH^2}[1-\cos({\frac{eH}{mc}t)}]$

$z=0$,

given that $\mathbf{r}=\mathbf{r}'=\mathbf{0}$ at $t=0$.

I started

$\frac{d\mathbf{r}}{dt} \wedge H=\frac{dy}{dt}\mathbf{i}-\frac{dx}{dt}\mathbf{j}$

Then I wrote the $3$ systems of equation(I am not too sure of this)

$m\frac{d^2r}{dt^2}=\frac{e}{c}\frac{dx}{dt}H$

$m\frac{d^2r}{dt^2}=eEy-\frac{e}{c}\frac{dy}{dt}H$

$m\frac{d^2r}{dt^2}=0$

Then, I presume I should ise the method of ordinary differential equation. But I do not know where to start.

Answer 1

Component form of the problem is

$x''=\frac{He}{mc}y'$ $(eqn$ $ 1)$

$y''=\frac{eE}{m}-\frac{He}{mc}x'$ $(eqn$ $2)$

$z''=0$ $(eqn$ $3)$

$\mathbf{From}$ $\mathbf{eqn}$ $\mathbf{3}$

$z'=A \rightarrow z''=At+k$

Since $z'=0 \rightarrow A=0$ and $z=0 \rightarrow k=0$

$\therefore z=0$

---

$\mathbf{From}$ $\mathbf{eqn}$ $\mathbf{1}$ $\mathbf{and}$ $\mathbf{2}$

Let's eliminate $y$:

$eqn$ $(1)$ $\mathbf{X}$ $D$: $D^3x=D^2\frac{eH}{mc}y$

$eqn$ $(2)$ $\mathbf{X}$ $(\frac{eH}{cm})$: $\frac{eE}{cm}D^2y=\frac{e^2EH}{cm^2}D\frac{eH}{mc}x$

Solving

$D^3 x=\frac{e^2EH}{cm^2}-(\frac{He}{mc})^2Dx$

Particular solution:

$D[D^2+(\frac{He}{cm})^2x]=\frac{e^2EH}{cm^2}$

$x_p=\frac{Ec}{H}t$

Auxilliary equation:

$m[m^2+(\frac{He}{cm})^2]=0 \rightarrow x_c=C_1+C_2 \cos{(\frac{He}{cm})t}+C_3 \sin{(\frac{He}{cm})t}$

$\therefore x=C_1+C_2 \cos{(\frac{He}{cm})t}+C_3 \sin{(\frac{He}{cm})t}+\frac{Ec}{H}$

---

Having obtained $x$,

$x'=-\frac{He}{cm} C_2 \sin{(\frac{He}{cm})t}+\frac{He}{cm} C_3 \cos{(\frac{He}{cm})t}+\frac{Ec}{H}$

$x''=-(\frac{He}{cm})^2C_2 \cos{(\frac{He}{cm})t}-(\frac{He}{cm})^2 C_3 \sin{(\frac{He}{cm})t}$

From $eqn$ $2$,

$y'=\frac{cm}{He}x''$

$y'=-\frac{cm}{He}[C_2 \cos{(\frac{He}{cm})t}+C_3 \sin{(\frac{He}{cm})t}]$

$y=-C_2 \sin{(\frac{He}{cm})t}+C_3\cos{(\frac{He}{cm})t}+C_4$

Substituting initial values.

$x=0 \rightarrow 0=C_1+C_2$

$x'=0 \rightarrow 0=\frac{He}{cm}C_3+\frac{Ec}{H}$

$\therefore C_3=-\frac{Ec^2m}{eH^2}$

$y'=0 \rightarrow 0=-\frac{cm}{He}C_2$

$C_2=0; C_1=0$

$y=0 \rightarrow 0=C_3+C_4 \therefore C_4=-C_3$

---

$x=\frac{Ec}{H}t-\frac{Ec^2m}{eH^2} \sin{(\frac{He}{cm})}t$

$y=\frac{Ec^2m}{eH^2} [1-\cos{(\frac{He}{cm})}t]$

$z=0$

Answer 2

Your three differential equations should be the individual components of $\underline{\ddot{r}}$

So you should have $$m\ddot{x}=\frac{eH}{c}\dot{y}$$ $$m\ddot{y}=eE-\frac{eH}{c}\dot{x}$$ And $$m\ddot{z}=0$$

I hope this helps.