# Differential equations & integration text problems

• Nov 12th 2007, 12:26 PM
Instigator
Differential equations & integration text problems
I need help with two problems:

The equation for the motion for a piece of metal is:

dv/dt = g-4v

find v and show that there is a limiting velocity

and: when is the piece of metal falling at g/10 m/s

Problem #2:

dV/dt = k(Vo - V) is the rate of evaporating water

in 20 days, 50% of the water evaporates; how much water (in percent) is remaining after 50 days?

Thanks for your help. It would be nice if you could give me a very easy to understand explanation of how this works. But if you can't, merely a solution is also helpful. Thanks for your support.
• Nov 12th 2007, 08:01 PM
topsquark
Quote:

Originally Posted by Instigator
The equation for the motion for a piece of metal is:

dv/dt = g-4v

find v and show that there is a limiting velocity

and: when is the piece of metal falling at g/10 m/s

$\frac{dv}{dt} = g - 4v$

$\frac{dv}{dt} + 4v = g$

Solve this how you like, either with integrating factors or by using the homogeneous and particular solutions. I get:
$v(t) = Ae^{-4t} + \frac{g}{4}$
where A is a constant determined by initial conditions.

The question is asking, what is
$\lim_{t \to \infty} v(t)$

So
$\lim_{t \to \infty} (Ae^{-4t} + \frac{g}{4} )$

$= \frac{g}{4}$

When is the object moving at $v = \frac{g}{10}$?
For this we need a value for A. It is sensible to choose $v(0) = 0$, so $A = -\frac{g}{4}$

Thus
$v(t) = -\frac{g}{4}e^{-4t} + \frac{g}{4}$

So when $v(t) = \frac{g}{10}$:
$\frac{g}{10} = -\frac{g}{4}e^{-4t} + \frac{g}{4}$

$\frac{1}{10} = -\frac{1}{4}e^{-4t} + \frac{1}{4}$

$\frac{1}{4}e^{-4t} = -\frac{1}{10} + \frac{1}{4}$

$e^{-4t} = \frac{3}{5}$

$-4t = ln \left ( \frac{3}{5} \right )$

$t = -\frac{1}{4} \cdot ln \left ( \frac{3}{5} \right )$

This is approximately at t = 0.127706 s.

-Dan
• Nov 13th 2007, 12:56 PM
Instigator
Thanks a lot, but can you maybe also tell me how exactly you can use integration to get drom dv/dt to v ? That would be very nice.
• Nov 13th 2007, 01:55 PM
topsquark
Quote:

Originally Posted by Instigator
Thanks a lot, but can you maybe also tell me how exactly you can use integration to get drom dv/dt to v ? That would be very nice.

Sure.

$\frac{dv}{dt} = g - 4v$

$\frac{dv}{dt} + 4v = g$

The most basic method (which is what I assume you want) is the method of finding an integrating factor.

So what can we multiply the LHS of the equation by to make it the derivative of a product?

Well:
$IF(t) = e^{\int 4~dt} = e^{4t}$

So
$e^{4t} \cdot \left ( \frac{dv}{dt} \right ) + 4ve^{4t} = ge^{4t}$

$\frac{d}{dt}(ve^{4t} ) = ge^{4t}$

$ve^{4t} = \int ge^{4t}~dt$

$ve^{4t} = g \cdot \frac{1}{4}e^{4t} + A$
(A is the arbitrary constant.)

$ve^{4t} = \frac{g}{4} \cdot e^{4t} + A$

$v = \frac{g}{4} + Ae^{-4t}$

(If you are more advanced than this, then you might note that the original differential equation is linear and has constant coefficients. So you can solve the homogeneous version of the equation and use variation of parameters to find the particular solution.)

-Dan