# 1st Order Differential; bit confused.

1. Feb 6, 2009

### Gregg

1. The problem statement, all variables and given/known data

A raindrop falls from rest through mist. Its velocity vms−1 vertically downward, at time t seconds after it starts to fall is modelled by the differential equation

$(1+t)\frac{dv}{dt} + 3v = (1+t)g-6$

Solve the differential equation to show that

$v=\frac{g}{4}(1+t)-2+(2-\frac{g}{4})(1+t)^{-3}$

2. Relevant equations

3. The attempt at a solution

$(1+t)\frac{dv}{dt} + 3v = (1+t)g-6$

$\frac{dv}{dt} + \frac{3v}{1+t} = g-\frac{6}{(1+t)}$

Let I be the integration factor

$I = e^{\int{p(x)}dx}$
$p(x)=\frac{3}{1+t}$
$I = (1+t)^3$

$(1+t)^3\frac{dv}{dt} + 3v (1+t)^2 = g(1+t)^3-6(1+t)^2$

$(1+t)^3 v = \int g(1+t)^3-6(1+t)^2 dt$

$(1+t)^3 v = \frac{g}{4} (1+t)^4 - \frac{6}{3}(1+t)^3 + c$

$v = \frac{g}{4} (1+t) - 2 + \frac{c}{(1+t)^3}$

Why is $c=2-\frac{g}{4}$

EDIT: don't worry now, it's the value of V when t = 0. Oops !

2. Feb 6, 2009

### rootX

Assuming: all is correct.

Use this initial condition
v(0) = 0
and substitute into your final equation.

Edit: Didn't see the edit