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I need to to model a raindrop's velocity as it is falling with respect to time.

The assumptions made are that air resistance is negligible and that the raindrop is spherical

I was able to to calculate and solve the differential equation for the change in radius over time to be: $$ r(t) = \frac{k}{p}t + r(0) $$ where $k$ is a proportionality constant, $p$ is the density of the raindrop and $r(0)$ is the initial radius.

Where I am stuck is where the differential equation for velocity of the raindrop is given but I need to solve the equation with the integrating factor method. Given that $r(0) = 3$ $$ \frac{dv}{dt} + \frac{3r'}{r}v = g $$ where $r = r(t)$ as above and $g$ = gravitational force constant.

The solution for this DE should be: $$ v(t) = \frac{pg}{4k} r + \frac{C}{r^3} $$ where $C$ is the integrating constant.

I am unsure of how I would calculate the integrating factor when there is a $r$ and $r'$ in the integral as well as how I should use this integrating factor to solve the differential equation.

I was given a hint that the chain rule would help with saving some work

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    $\begingroup$ Typeset mathematical terms using MathJax. Here's the tutorial. $\endgroup$ – SarGe Sep 2 '20 at 8:35
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The obvious integrating factor is (choosing convenient integration constants on-the-fly) $$ \exp\left(3\int\frac{r'}rdt\right)\overset{\text{chain rule}}=\exp\left(3\int\frac{dr}r\right)=\exp(3\ln r)=r^3, $$ so that then $$ (r^3v)'=r^2(rv'+3r'v)=gr^3 $$ Now it remains to find the integral of $r(t)^3=(r_0+r_1t)^3$ which is easy to do in this case. $$ r^3v=\frac{gr^4}{4r_1}+C $$

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  • $\begingroup$ Thank you so much. This clears a lot of things up for me. $\endgroup$ – Nikki Sep 2 '20 at 9:23

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