I made my own air cannon at home and would like to determine its launch velocity. I worked out the following differential equation:

$$\frac{d^2}{dt^2} x(t) = \frac{A}{m} \left(\frac{P_0V_0}{V_0 + Ax(t)} - P_\text{atm}\right) $$

How would I solve this equation?

Can anyone verify whether or not this is a solution?

$$x(t) = \frac{V_0}{A}\left\{1 - \exp\left[\text{erfi}^{-1}\left( At \sqrt{\frac{2P_0}{\pi V_0m}}\right)^2 \right] \right\}$$

  • $\begingroup$ can you say something to the variables? $\endgroup$ – Dr. Sonnhard Graubner Oct 14 '15 at 17:02
  • $\begingroup$ what does Wolfram alpha say? $\endgroup$ – Dr. Sonnhard Graubner Oct 14 '15 at 17:05
  • $\begingroup$ Wolfram alpha cannot solve it. $\endgroup$ – Halbort Oct 14 '15 at 19:03

Getting rid of strange constants, your differential equation can be written as $$ x''=\frac{c_1}{c_2+x}+c_3. $$ Multiply by $x'$ and integrate, to get $$ \frac{1}{2}(x')^2=c_1\log(c_2+x)+c_3x+c_4 $$ Separating variables, you need to calculate an integral of the form $$ \int \frac{1}{\sqrt{2c_1\log(c_2+x)+2c_3x+2c_4}}\,dx. $$ I suggest that you use your air cannon for that purpose.

  • 1
    $\begingroup$ In fact, you need to calculate and invert such an integral. That second part is likely to be about as hard as the first part. Also, because numerics will be required, it might help to specify the problem as fully as possible: if $x(0)=x_0,x'(0)=v_0$, then you have $1/2 v_0^2=c_1 \log(c_2)+c_3 x_0 + c_4$. So you can solve for $c_4$. Having done that, you need to solve $t=\int_{x_0}^{x(t)} (2 c_1 \log(c_2+x) + 2c_3 x + 2 c_4)^{-1/2} dx$ for $x(t)$. $\endgroup$ – Ian Oct 14 '15 at 19:27

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.