# Is there a function whose derivative is proportional to that function squared?

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 $$v(t)$$?

• you get this $$\frac 1 p=2\frac {b^2}{m^2}t+K$$ – Isham Oct 14 '18 at 17:19

## 1 Answer

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

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

• 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. – Null Spark Oct 14 '18 at 17:33
• @NullSpark just elementary fractions nothing to do with differential calculus ..I will add some lines – Isham Oct 14 '18 at 17:33
• @NullSpark is it more clear now ? – Isham Oct 14 '18 at 17:38
• Yes. Thank you very much. – Null Spark Oct 14 '18 at 17:40
• Thanks. I thought that I would tidy up in case it confused anyone else. – badjohn Oct 14 '18 at 20:10