Integrating with respect to time a double derivative $\ddot\phi + \frac bm\dot\phi = \frac{F}{mr}$ $$\ddot\phi + \frac bm\dot\phi = \frac{F}{mr}$$
I want to integrate this to get an equation for $\dot\phi(t)$ but I don't know how to integrate double derivatives. The answers say that the homogeneous equation associated with it is:
$$\ddot\phi + \frac bm\dot\phi = 0$$
and by integrating that you get 
$$\dot\phi(t) = A \exp(−bt/m)$$
How do they integrate this homogeneous equation to get $A \exp(−bt/m)$?
I am only familiar with integrating first derivatives. Also if it gives $A \exp(−bt/m)$, then the non homogeneous equation is according to the solutions that 
$$\ddot\phi + \frac bm\dot\phi = \frac F{mr}$$
integrates to give 
$$\dot\phi(t) = A\exp(−bt/m) + \frac F{br}.$$
I also don't get this, because wouldn't the non homogeneous equation give $A\exp(−bt/m) + F/mr$ instead of $F/br$??
 A: A general way to solve equations on the form
$$\ddot{\phi} + f(t)\dot{\phi} = g(t)$$
is to multiply by the so-called integrating factor $e^{\int f(t) dt}$. Since $\frac{d}{dt}\left(\dot{\phi}e^{\int f(t) dt}\right) = e^{\int f(t) dt}(\ddot{\phi} + f(t) \dot{\phi})$ then if we multiply the equation with $e^{\int f(t) dt}$ we get
$$\frac{d}{dt}\left(\dot{\phi}e^{\int f(t) dt}\right) = g(t)e^{\int f(t) dt}$$
which we can now integrate directly to get
$$\dot{\phi} = e^{-\int f(t) dt}\int g(t)e^{\int f(t) dt}dt$$
For your case $g(t) = \frac{F}{mr}$ and $f(t) = \frac{b}{m}$ so $e^{\int f(t) dt} = e^{\frac{bt}{m}}$. If you perform the integration (don't forget the integration constant) you'll get the desired answer.
A: EDIT: Sorry, my answer is wrong of course..
The trick to answering how to integrate the double derivative is to treat them as quantity of their own, not as a derivative of some other quantity. If you express:
$$\ddot\phi = \frac{d\dot\phi}{dt}$$
you have $$\frac{d \dot\phi}{dt} + \frac bm\dot\phi = \frac{F}{mr}$$
Now just solve this for $\dot\phi$, forgetting that it is derivative of some quantity. The solution should read
$$\dot\phi(t) = A\exp(−bt/m) + \frac F{mr}.$$
as you pointed out, I believe that there is a typo in your solutions.
A: You don't have to integrate $x'(t)$ you can for instance take $y(t) = x'(t) $ then you get the homogenous equation 
$y'(t) + (b/m)y(t) = 0 $
Which you know how to solve ,then solve for $ y(t)$ ,but $y(t) = x'(t) $ so you have actually solved for $x'(t)$
Also I think this whole m and b thing you mentioned in the may be just a typo (I didnt look much into it, sorry)
