# Solve the equations for v x </msub> and v y </msub> :

Solve the equations for ${v}_{x}$ and ${v}_{y}$ :
$m\frac{d\left({v}_{x}\right)}{dt}=q{v}_{y}B\phantom{\rule{2em}{0ex}}m\frac{d\left({v}_{y}\right)}{dt}=-q{v}_{x}B$
by differentiating them with respect to time to obtain two equations of the form:
$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha }^{2}u=0$
where $u={v}_{x}$ or ${v}_{y}$ and ${\alpha }^{2}=qB/m$. Then show that $u=C\mathrm{cos}\alpha t$ and $u=D\mathrm{sin}\alpha t$, where C and D are constants, satisfy this equation
Whenever I differentiate the first equation with respect to time, I get a resulting equation with the form:
$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha }^{2}\frac{du}{dt}=0$
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cegielnikmzjkf
Differentiating the first equation -
$m{\stackrel{¨}{v}}_{x}=q{\stackrel{˙}{v}}_{y}B\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}m{\stackrel{¨}{v}}_{x}=-\frac{{q}^{2}{B}^{2}}{m}{v}_{x}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\stackrel{¨}{v}}_{x}+\left(\frac{qB}{m}{\right)}^{2}{v}_{x}=0$
Where I have substituted for ${\stackrel{˙}{v}}_{y}$ from equation 2.
You will get a similar equation on differentiation equation 2 for ${\stackrel{¨}{v}}_{y}$. And then of course given $\alpha =\frac{qB}{m}$ you can get solutions as -
$u=A\mathrm{cos}\left(\alpha t\right)+B\mathrm{sin}\left(\alpha t\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}u={v}_{x},{v}_{y}$