2
$\begingroup$

sorry I'm having some trouble evaluating this integral

$\frac{dv}{dt} = -k(v-gt)^2-g$ where g and k are constants

I'm assuming you just separate and integrate but I cannot seem to get it to work out.

$\endgroup$
  • $\begingroup$ What part of "separate and integrate" didn't work out, aside from the grammatical disparity? $\endgroup$ – abiessu Oct 16 '14 at 4:56
  • $\begingroup$ I'm guessing the "separate" part, given that it's not separable... $\endgroup$ – Daniel McLaury Oct 16 '14 at 5:06
3
$\begingroup$

Let $y=v-gt$. Then $\frac{dv}{dt}=\frac{dy}{dt}+g$. So our differential equation can be rewritten as $$\frac{dy}{dt}+g=-ky^2-g,$$ and then as $$\frac{dy}{dt}=-(ky^2+2g).$$ This is a separable differential equation. We are solving $$\frac{dy}{ky^2+2g}=-dt.$$ Integrate. We will get an arctan on the left.

$\endgroup$

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.