0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.