# Problem simplifying the solution of an ODE.

1. Feb 21, 2013

### -Dragoon-

So, I was following the derivation in my physics book of:
$x(t) = c_1e^-(\frac{\gamma t}{2})\cos(\omega_d t)+c_2e^-(\frac{\gamma t}{2})\sin(\omega_d t)$

Until they simply get to this in one step:
$Ae^-(\frac{\gamma t}{2})\cos(\omega_d t + \phi)$

I've tried reading many other sources for this derivation of the underdamped oscillator, and I follow up until this last critical step and they don't hint at the omitted steps.

2. Feb 21, 2013

### HallsofIvy

Staff Emeritus
That's a pretty standard step. Use the identity cos(A+ B)= cos(A)cos(B)- sin(A)sin(B).

Comparing that to $\alpha cos(\theta)+ \beta sin(\theta)$
($\alpha= c_1e^{-\gamma t/2}$, $\beta= c_2e^{-\gamma t/2}$ and $\theta= \omega_d t$.)
We need $cos(A)= \alpha$ and $sin(A)= \beta$. Of course, that is not possible unless $\alpha^2+ \beta^2= 1$. If that is not true, then we multiply and divide by $\alpha^2+ \beta^2$:
$$(\alpha^2+ \beta^2)\left(\frac{\alpha}{\alpha^2+ \beta^2}cos(\theta)+ \frac{\beta}{\alpha^2+ \beta^2}sin(\theta)\right)$$

3. Feb 21, 2013

### LCKurtz

I think Halls meant$$\sqrt{\alpha^2+ \beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+ \beta^2}}cos(\theta)+ \frac{\beta}{\sqrt{\alpha^2+ \beta^2}}sin(\theta)\right)$$

4. Feb 21, 2013

### -Dragoon-

Got it, thanks everyone.