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I've just begun to study ODE and I have to solve this set of equations:

$$\frac {d v_x}{dt}=\omega v_y$$ $$\frac {d v_y}{dt}=-\omega v_x$$

I have made these steps:

$$v_y=\frac{1}{\omega} \frac{d v_x}{dt}$$ $$ \frac{d}{dt}(\frac{1}{\omega} \frac{dv_x}{dt})=-\omega v_x$$

then

$$\frac{1}{\omega}\frac{d^2v_x}{dt^2}=-\omega v_x$$

and so $$\frac{d^2v_x}{dt^2}=-\omega^2 v_x$$

If I'm not wrong, the solution of this DE is $$\alpha cos (\omega t) +\beta sin (\omega t)$$

Starting from this solution, how can I obtain something like $Acos(\omega t + \phi)$?

Thanks a lot!

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1 Answer

up vote 1 down vote accepted

$$\alpha \cos (\omega t) +\beta \sin (\omega t)$$

$$=\sqrt{\alpha^2+\beta^2} ( \frac{\alpha}{\sqrt{\alpha^2+\beta^2}}\cos (\omega t)+\frac{\beta}{\sqrt{\alpha^2+\beta^2}}\sin (\omega t)) $$

Take $\sin \phi=\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}$ and $\cos \phi=\frac{\beta}{\sqrt{\alpha^2+\beta^2}}$

$\Rightarrow \tan \phi=\frac{\alpha}{\beta}\Rightarrow\phi=\arctan\frac{\alpha}{\beta}$

Take $A=\sqrt{\alpha^2+\beta^2}$

Then we have, $$\alpha \cos (\omega t) +\beta \sin (\omega t)=A\cos(\omega t+\phi)$$

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@sunrise I think you wanted this explanation. –  Abhra Abir Kundu Feb 11 '13 at 13:27
    
Wonderful!!!! Thank you so much!! –  sunrise Feb 11 '13 at 15:12
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