Solutions to SHMO equation in Feynman's Thesis I'm reading Feynman's Thesis, and have some background in math and physics, but I'm not sure where Feynman gets his solution to his harmonic oscillator equation.  He gives three different formulations.  The first includes an integral.  Is this standard Fourier Analysis stuff?  Can you give me some kind of link? 
One of them is
$$ x(t) = x(0) \cos \omega t + \dot{x}(0) \frac {\sin\omega t}{\omega} + \frac {1}{m \omega}\int_o^t \gamma (s) \sin \omega (t-s) ds $$  
$$\gamma = I_y + I_z $$.  The "I" is a function involving the coordinates of positions y and z.  
$$  m\ddot{x} + m\omega^2 x = [ I_y + I_z ]  $$
 A: This comes from linearity, the solution of the homogenous equation, plus the response to a delta-function kick.
The homogenous equation gives the first two terms. The equation
$$ \ddot{x} + \omega^2 x = 0 $$
is solved by a combination of cos and sin which reflect the initial position and velocity
$$ x(t) = x_0 cos(\omega t) + {v_0\over \omega} sin(\omega t) $$
You can verify that this is correct by looking near t=0 to see that it starts at $x_0$ with velocity $v_0$.
Now consider adding a delta-function kick at some time $t_0$
$$ \ddot{x} + \omega^2 x = \delta (t-t_0)$$
You want the solution to this equation which only has influence into the future, meaning that for $t<t_0$, $x(t)=0$. The delta function is an impulsive kick which makes the particle move with a velocity 1 at time $t_0+\epsilon$, so that the solution from this point onward is the solution to the equation with initial velocity 1 starting at $t=t_0$, or
$$ x(t) = {1\over \omega} sin(\omega (t-t_0))$$
Now you consider the source term $\gamma(t)$ to be a sum of delta functions at each time, each one producing this response. This leads to a total response (by linearity) of
$$ x(t) = \int_0^t{\gamma(t')\over \omega} sin(\omega(t-t')) dt' $$
And this integral is the last term in Feynman's solution. The forward effects of the sin kernel means that this solution doesn't affect the initial conditions, so that you just add this to the solution with the given initial conditions to find the general solution.
The only things he uses here are linearity of the equation, plus the solution of the homogenous equation for a given initial condition.
