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I am working on a Physics problem. The problem involves a ball of mass m falling through the air, with the drag force at a time t being equal to $bv^3(t)$. The problem wants me to find and solve a differential equation for the instantaneous velocity at a given time. I did some research, and I found an article on the Oregon State University website which explains how to solve the equation assuming that drag is directly proportional to velocity. I tried solving it the same way, and I eventually end up at a point where an auxiliary function $p'(t) = -2\frac{b^2}{m^2}p^2(t)$. This is pretty much where the problem dead-ended for me, though.

My two questions are:

1) Is there an elementary function / combination of elementary functions such that $f'(t)\propto f^2(t)$?

2) How do I actually solve this equation for $v(t)$?

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    $\begingroup$ you get this $$\frac 1 p=2\frac {b^2}{m^2}t+K$$ $\endgroup$ – Isham Oct 14 '18 at 17:19
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1) Is there an elementary function / combination of elementary functions such that $f′(t)∝f^2(t)$ ?

Yes. Take for example $f(x)=1/x$ we have $f'(x)=-\frac 1 {x^2}$ and also $f^2(x)=\frac 1 {x^2}$ $$p'(t) = -2\frac{b^2}{m^2}p^2(t)$$ This differential equation is separable. If you solve the equation in p you get : $$\int \frac {dp}{p^2}= -2\int \frac{b^2}{m^2}dt$$

$$\frac 1 p=2\frac {b^2}{m^2}t+K$$ Invert fractions on both sides you get $$ p(t)=\frac 1{ {2\frac {b^2}{m^2}t+K}}$$ multiply denominator and numerator by $m^2$ $$ p(t)=\frac {m^2} { {2 {b^2}t+Km^2}}$$ Substitute $C=m^2K$ which is just a constant

$$\implies p(t)=\frac {m^2} {2{b^2}t+C}$$

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  • $\begingroup$ Thank you very much. I was wondering if you could explain your last step, though? I don't understand how you get that m^2 into the numerator. $\endgroup$ – Null Spark Oct 14 '18 at 17:33
  • $\begingroup$ @NullSpark just elementary fractions nothing to do with differential calculus ..I will add some lines $\endgroup$ – Isham Oct 14 '18 at 17:33
  • $\begingroup$ @NullSpark is it more clear now ? $\endgroup$ – Isham Oct 14 '18 at 17:38
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    $\begingroup$ Yes. Thank you very much. $\endgroup$ – Null Spark Oct 14 '18 at 17:40
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    $\begingroup$ Thanks. I thought that I would tidy up in case it confused anyone else. $\endgroup$ – badjohn Oct 14 '18 at 20:10

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