How to solve this differential equation:  $$ m*(dv/dt)=-m g- r v$$  where m, r and g are constants.  I am trying to rewrite the equation as dv=dt form, but I do not know how to do it. I get the solution from the mathematica as   $$ (mg/r+v)*e^{(-rt/m)}-(mg/r)$$
I've seen many variants of this problem online, but not quite the same as this, so I don't believe this is a duplicate.  The famous differential equation problem models a skydiver jumping out of a plane as  $$m \frac{dv}{dt} = -mg-cv$$ where m is the mass of the skydiver, g is acceleration due to gravity (9.81 m/s) and c is the coefficient of air resistance.  Here is my question:     Suppose the skydiver exits the stationary helicopter at an altitude of 2000m and opens the parachute at 500m. For this exercise, the parachute opens instantaneously. Suppose m= 70kgs, the coefficient of air resistance is c = 14 for a free falling body and c = 105 after the parachute opens. When does the skydiver open the parachute? When does the Skydiver hit the ground?   So my attempt at a solution kinda starts like this, but this is where I get stuck:  Let's assume that s(t) is the distance the body falls in time t from its initial point of release, then $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$. Thus we have a second order differential equation: $m \frac{d^2s}{dt^2} = -mg-c\frac{ds}{dt}$. By simply manipulation, we get $m \frac{d^2s}{dt^2} + c\frac{ds}{dt} = mg$. I originally started thinking of finding the integrating factor, however, this seems more complicated and won't help produce the equation that I need.  Would anyone be able to help? Much thanks in advanced!
How to create a new partition from the existing partition? I wanna create a new partition from the existing root partition.My root partition is about 250GB.I wanna create a new partition say 150 GB taking space from this root partition.Can anyone help me with the procedure on doing this?  output: fdisk -l  karthick@karthick:~$ sudo fdisk -l [sudo] password for karthick:   Disk /dev/sda: 250.0 GB, 250000000000 bytes 255 heads, 63 sectors/track, 30394 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x28219744  Device Boot      Start         End      Blocks   Id  System /dev/sda1   *           1          61      487424   83  Linux Partition 1 does not end on cylinder boundary. /dev/sda2              61       30395   243650561    5  Extended /dev/sda5              61         124      498688   82  Linux swap / Solaris /dev/sda6             124         186      498688   82  Linux swap / Solaris /dev/sda7             186       30395   242651136   83  Linux How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$? How can we compute the following limit:     $$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$   Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it. Controversy over subject-verb agreement in this sentence The sentence     Women driving cars is, of course, such a foreign sight to a society like   Saudi Arabia   The subject is not "women" (otherwise, the verb would have been 'are'); the subject, as I mean to use it, is the rarity of seeing women driving cars. The subject, in other words, is "women driving cars" as a thing. Does this give me licence to use 'is' here? Thank you. Exporting multi layered Illustrator to Photoshop Help with table multicolumn We all use mathematical induction to prove results, but is there a proof of mathematical induction itself? I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like   $$1+2+3+\cdots+n = \frac{n(n+1)}{2}$$  However.  I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.) Navigate folder structure in Open/Save File Dialog with keyboard? I want to enable GPU rendering, but there is no option in User Preferences > System:    Why is this? How can I get cycles to render using my GPU? I noticed that Synonymizer is only a bronze badge. Yet in order to obtain this badge you need to have 2500 reputation or more to even create tag synonyms. Then, on top of that you also need to be able to find a tag synonym.  Only 514 of these have been awarded... This makes the BRONZE badge Synonymizer rarer to have than most the gold medal badges. So my question is, why is this a bronze badge? How to install the most common plugins? Show that holomorphic $f_1, . . . , f_n $ are constant if $\sum_{k=1}^n \left| f_k(z) \right|$ is constant. Should I use a canonical link which refers to the current URL with pagination? Limiting double to 3 decimal places Why do we need a Stack Overflow in Portuguese? I have noticed there is a Stack Overflow in language differs from English which is global and mostly used by programmers.  I think there is no need for a site in other languages. Otherwise we will end up with 127 StackOverflow sites. I am afraid that we will lose the amount of people sharing data. How are parentheses interpreted at the command line? Can a Rogue (D&D 3.5) use sneak attack on both the first and second attack? I know this seems like a question you can simply Google but the answers vary and my clan is under the impression that I can only use it on my first attack (once per round) and not on my second attack.  From Googling, I've understood that as long as the monster / enemy is flat-footed, flanked or caught off-guard, I can apply sneak attack on both my first and second attack in a round.  Is this correct?  Is it normal if the clan simply decides that that gives me an unfair advantage in combat and gives me only one sneak attack per round? Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:  \begin{equation} I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt \end{equation}  For $n \in \mathbb{R},\:n \gt 1$  The method I took was to take the following integral:  \begin{equation}  \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \end{equation}  Where: $0 \leq k \lt n$. Here let $m = 1$.   with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the ), i.e.   \begin{align} \lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &amp;= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\ \lim_{k \rightarrow 0+}  \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt  &amp;= \lim_{k \rightarrow 0+}  \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\ \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&amp;= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\ &amp;=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right) \end{align}  Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:  \begin{equation}  J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right] \end{equation}  And in fact we may generalise:  \begin{equation}     \int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right] \end{equation}  Where $p \in \mathbb{N}$  This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used? How to bond/connect wires to neodymium magnets? After reading , I learned it is not advisable to solder a wire to neodymium magnets.  After more reading I found  post describing the use of conductive epoxy to bond electrical wires to neodymium magnets, a very messy process which takes several hours of waiting.  I want to ask this community if they have or suggest better methods to bond/connect a wire to a neodymium magnet to act as a conductive connection? How to find all the subclasses of a class given its name? I read it in Reader's Digest in the seventies. Archaeologists interpreted everyday objects as the TV set and the toilet seat to be sacred and precious.