Integrating with respect to time a double derivative $\ddot\phi + \frac bm\dot\phi = \frac{F}{mr}$

Question

$$\ddot\phi + \frac bm\dot\phi = \frac{F}{mr}$$

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 $F/br$??

Shouldn't there also be a factor of $t$ in $F/mr$? Otherwise, $\frac{d}{dt}(F/mr)=0$ (assuming that $F,,m,,r$ are all constants) — Kyle Kanos, Sep 03 '14 at 19:01

score 7 · Accepted Answer · answered Sep 03 '14 at 19:32

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.

score 1 · Answer 2 · answered Sep 03 '14 at 18:56

1

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.

answered Sep 03 '14 at 18:56

miha priimek

263

No it's not a typo, it really should be $\frac{F}{br}$. – Winther Sep 03 '14 at 20:05
Ahh of course, you're absolutely correct. – miha priimek Sep 03 '14 at 20:23
btw apart from $\frac{F}{mr} \to \frac{F}{br}$ then what you have written seems correct so just fix the typo and it should be fine. – Winther Sep 03 '14 at 20:29
Well, it seems correct, but I really did forget about the integrating factor... – miha priimek Sep 04 '14 at 05:07
@Winther i dont doubt the solutions but how did you figure it out to be F/br ? – user135688 Sep 04 '14 at 15:09
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See it for yourself - differentiate the \dot\phi with respect to time and plug it in the equation. Be careful to see the factor before the \phi\dot, I forgot about it.. – miha priimek Sep 04 '14 at 15:16

score 0 · Answer 3 · answered Sep 03 '14 at 18:50

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)

Integrating with respect to time a double derivative $\ddot\phi + \frac bm\dot\phi = \frac{F}{mr}$

3 Answers3