# Separation of Variables

• Jan 3rd 2010, 04:20 PM
junkedly
Separation of Variables
i need help with a problem involving separation of variables. here's the entire problem:

let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t > or = 0. after her parachute opens, her velocity satisfies the differential equation dv/dt = -2v - 32, with initial condition v(0) = -50

use separation of variables to find an expression for v in terms of t, where t is measured in seconds

anyone mind helping out?
• Jan 3rd 2010, 04:28 PM
dedust
Quote:

Originally Posted by junkedly
i need help with a problem involving separation of variables. here's the entire problem:

let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t > or = 0. after her parachute opens, her velocity satisfies the differential equation dv/dt = -2v - 32, with initial condition v(0) = -50

use separation of variables to find an expression for v in terms of t, where t is measured in seconds

anyone mind helping out?

hi jun

$\frac{dv}{dt} = -2v - 32$, then

$\frac{dv}{-2v - 32} = dt$

solve this differential equation
• Jan 3rd 2010, 04:30 PM
pickslides
$\frac{dv}{dt} = -2v - 32$

$\frac{dv}{-2v - 32} = dt$

Now integrate both sides.
• Jan 3rd 2010, 04:40 PM
junkedly
thanks for the quick replies

i had that much so far, but i when i integrate dv/-2v-32 i get an antiderivative that is a nonreal answer

i end up with -2ln(-2v - 32) = t, but then the initial condition doesn't work because a natural logarithm can't equal zero

thanks again for the help
• Jan 3rd 2010, 04:48 PM
skeeter
$\frac{dv}{-2v - 32} = dt$

$\frac{dv}{-2(v + 16)} = dt$

$\frac{dv}{v + 16} = -2 \, dt$

$\ln|v+16| = -2t+C$

finish ...
• Jan 3rd 2010, 04:51 PM
junkedly
a huuuuuge thank you skeeter

i always manage to overlook possible factoring

you saved me :)
• Jan 3rd 2010, 04:59 PM
junkedly
so adding the initial conditions of v(0) = -50 i get

ln|-50+16| = -2(0) + C

C = 3.526

therefore ln|v+16| = -2t + 3.526

does this look correct?
• Jan 3rd 2010, 05:25 PM
skeeter
Quote:

Originally Posted by junkedly
so adding the initial conditions of v(0) = -50 i get

ln|-50+16| = -2(0) + C

C = 3.526

therefore ln|v+16| = -2t + 3.526

does this look correct?

$\ln|v+16| = -2t+C$

$v+16 = e^{-2t+C} = e^{-2t} \cdot e^C$ ... let $e^C = A$

$v+16 = Ae^{-2t}$

$v = -16+Ae^{-2t}$

$v(0) = -50$ ...

finish