# differential equation help >.<

• April 27th 2008, 09:39 AM
tasukete
differential equation help >.<
hi all,

Im having trouble solving this differential equation, ive tried many times but just cant seem to get my algebra correct >.< and hence end up getting stuck.

$

dv/dt = -g - bv/m

$

pls help me
• April 27th 2008, 09:40 AM
Mathstud28
Quote:

Originally Posted by tasukete
hi all,

Im having trouble solving this differential equation, ive tried many times but just cant seem to get my algebra correct >.< and hence end up getting stuck.

$

dv/dt = /int -g - bv/m

$

pls help me

I am unsure of what this means....you have this $\frac{dv}{dt}=\int{g-\frac{bv}{m}}d?$

What is v function of and are the other variables constants?
• April 27th 2008, 09:44 AM
tasukete
Quote:

Originally Posted by Mathstud28
I am unsure of what this means....you have this $\frac{dv}{dt}=\int{g-\frac{bv}{m}}d?$

What is v function of and are the other variables constants?

sry i typed it incorrectly. its:

$
\frac{dv}{dt}= -g -\frac{bv}{m}
$

sry for the confusion
• April 27th 2008, 09:45 AM
tasukete
this is actually a physics question where b is the drag coefficient, g is gravity, m is mass, v is velocity
• April 27th 2008, 09:46 AM
tasukete
im asked to find the velocity, v, in terms of t,b,g,m
• April 27th 2008, 09:47 AM
Mathstud28
Quote:

Originally Posted by tasukete
sry i typed it incorrectly. its:

$
\frac{dv}{dt}= -g -\frac{bv}{m}
$

sry for the confusion

Ok this is how I see it since you have dt

$\frac{bv}{m}dv=-gdt$

integrating $\frac{bv^2}{2m}=-gt+C$

can you go from there?
• April 27th 2008, 09:48 AM
PaulRS
Quote:

Originally Posted by tasukete
sry i typed it incorrectly. its:

$
\frac{dv}{dt}= -g -\frac{bv}{m}
$

sry for the confusion

We have: $
\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}} = - 1 \Rightarrow \int_0^t {\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}}dt} = - t
$

Now let $u=v$ on the left hand side and it follows easily

$
\int_0^t {\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}}dt} = \int_{v\left( 0 \right)}^{v\left( t \right)} {\frac{{dv}}
{{g + \tfrac{{b \cdot v}}
{m}}}}
$

;)
• April 27th 2008, 09:51 AM
tasukete
Quote:

Originally Posted by PaulRS
We have: $
\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}} = - 1 \Rightarrow \int_0^t {\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}}dt} = - t
$

Now let $u=v$ on the left hand side and it follows easily

$
\int_0^t {\frac{{v'}}
{{g + \tfrac{{b \cdot v}}
{m}}}dt} = \int_{v\left( 0 \right)}^{v\left( t \right)} {\frac{{dv}}
{{g + \tfrac{{b \cdot v}}
{m}}}}
$

;)

gosh that looks very complicated.. (Headbang)
• April 27th 2008, 09:53 AM
PaulRS
Quote:

Originally Posted by tasukete
gosh that looks very complicated.. (Headbang)

That's not true, poor integral (Crying)

Note that $
\left[ {\ln \left( {g + \tfrac{{b \cdot v}}
{m}} \right)} \right]^\prime = \frac{{\tfrac{{b \cdot v'}}
{m}}}
{{g + \tfrac{{b \cdot v}}
{m}}}
$
(Wink)
• April 27th 2008, 10:08 AM
tasukete
Quote:

Originally Posted by Mathstud28
Ok this is how I see it since you have dt

$\frac{bv}{m}dv=-gdt$

integrating $\frac{bv^2}{2m}=-gt+C$

can you go from there?

im unsure what u just did there
• April 27th 2008, 10:29 AM
tasukete
can i ask for a confirmation. unsure whether this is correct direction to take..

$
\frac{dv}{dt} = -g - \frac{bv}{m}
$

$
\frac{dt}{dv} = \frac{-1}{g} - \frac{m}{bv}
$

$
t = \int \frac{-1}{g} - \frac{m}{bv}dv
$

$
t = \frac{-v}{g} - \frac{m}{b}logv + c
$
• April 27th 2008, 10:34 AM
Moo
Quote:

Originally Posted by tasukete
can i ask for a confirmation. unsure whether this is correct direction to take..

$
\frac{dv}{dt} = -g - \frac{bv}{m}
$

$
\frac{dt}{dv} = \frac{-1}{g} - \frac{m}{bv}
$

$
t = \int \frac{-1}{g} - \frac{m}{bv}dv
$

The inverse of $a+b$ is not $\frac 1a + \frac 1b$ ...
• April 27th 2008, 10:39 AM
tasukete
Quote:

Originally Posted by Moo
The inverse of $a+b$ is not $\frac 1a + \frac 1b$ ...

oh my...stupid me
• April 27th 2008, 10:57 AM
tasukete
$
\frac{dv}{dt} = -g - \frac{bv}{m}
$

$
\frac{dv}{dt} = \frac{-gm - bv}{m}
$

$
\frac{dt}{dv} = \frac{m}{-gm - bv}
$

$
t = \int \frac{m}{-gm - bv}dv
$

$
t = -m log(gm + bv) + c
$