# Solving a differential equation using Integrating Factor method

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

• Typeset mathematical terms using MathJax. Here's the tutorial. – SarGe Sep 2 '20 at 8:35

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. $$r^3v=\frac{gr^4}{4r_1}+C$$