I got the following formula: $$m*\frac{dv}{dt} = -k*v²$$ where m and k are constants.

These where my steps to solve it: $$\frac{dv}{dt} = \frac{-k*v²}{m}$$ $$m*dv=-k*v²dt$$ $$\frac{1}{-k*v²}dv = \frac{1}{m}dt$$ Integrate both sides: $\int\frac{1}{-k*v²}dv = \int\frac{1}{m}dt$ Since m &k are constants: $\frac{-1}{k}*\int\frac{1}{v²}dv = \frac{1}{m}*\int1*dt$ $$\frac{-1}{k}*\frac{-1}{v} + C_1= \frac{1}{m}*t + C_2$$ Let's make both C's one variable: C $$\frac{1}{kv}= \frac{t}{m}+ C$$ Now let's inverse both sides: $$kv = \frac{m}{t+cm}$$ $$v=\frac{m}{kt+cmk}$$ Wolframalpha gives this answer but without the k in cmk and with a - in front of the right side of the equation. When I fill the equation back into the original formula I also don't get the same answer. What did I do wrong here?

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    $\begingroup$ You do get the same as Wolframalpha, the $c_1$ there is simply $c_1=-kc$. $\endgroup$ – Mathematician 42 Mar 14 '17 at 10:15
  • $\begingroup$ And if $c$ is a constant $cmk$ is another one (which will be fixed by a condition). $\endgroup$ – Claude Leibovici Mar 14 '17 at 10:17
  • $\begingroup$ But why doesn't wolfram alpha show that? $\endgroup$ – Sam Liemburg Mar 14 '17 at 10:19
  • $\begingroup$ It simply separated the equation different then you did (it also moved $k$ to the left). Doing so slighty changes the look of the solution, but it's still the same thing. $\endgroup$ – Mathematician 42 Mar 14 '17 at 10:23
  • $\begingroup$ But shouldn't it atleast tell me what c_1 is? $\endgroup$ – Sam Liemburg Mar 14 '17 at 10:28

I always prefer not to use integration constants when solving differential equation (see confusion above). Rather, I would write the following

$$-\frac{1}{k}\int_{\nu(t_0)}^{\nu(t)}\frac{d\nu'}{(\nu')^2} = \frac{1}{m}\int_{t_0}^t dt'. $$

This leads to $$\nu(t) = \frac{m \nu(t_0)}{m + k \nu(t_0)(t-t_0)} $$

and there is no ambiguity whatsoever about ''choosing'' the constants of integration.

  • $\begingroup$ But can you define the limits like that? $\endgroup$ – Sam Liemburg Mar 14 '17 at 12:42
  • $\begingroup$ Yes. I am free to chose where I integrate over. If I choose to integrate $t'$ between $t_0$ and $t$, then we must integrate $\nu'$ between $\nu(t_0)$ and $\nu(t)$. You can do this without ambiguity since your differential equation tells you from the start that $\nu$ is injective... $\endgroup$ – Stefano Mar 14 '17 at 12:47

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