Solve the equations for ${v}_{x}$ and ${v}_{y}$ :

$m\frac{d({v}_{x})}{dt}=q{v}_{y}B\phantom{\rule{2em}{0ex}}m\frac{d({v}_{y})}{dt}=-q{v}_{x}B$

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:

$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha}^{2}\frac{du}{dt}=0$

$m\frac{d({v}_{x})}{dt}=q{v}_{y}B\phantom{\rule{2em}{0ex}}m\frac{d({v}_{y})}{dt}=-q{v}_{x}B$

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:

$\frac{{d}^{2}u}{d{t}^{2}}+{\alpha}^{2}\frac{du}{dt}=0$