# differential equation help >.<

• Apr 27th 2008, 08:39 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.

$\displaystyle dv/dt = -g - bv/m$

pls help me
• Apr 27th 2008, 08:40 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.

$\displaystyle dv/dt = /int -g - bv/m$

pls help me

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

What is v function of and are the other variables constants?
• Apr 27th 2008, 08:44 AM
tasukete
Quote:

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

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

sry i typed it incorrectly. its:

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

sry for the confusion
• Apr 27th 2008, 08:45 AM
tasukete
this is actually a physics question where b is the drag coefficient, g is gravity, m is mass, v is velocity
• Apr 27th 2008, 08:46 AM
tasukete
im asked to find the velocity, v, in terms of t,b,g,m
• Apr 27th 2008, 08:47 AM
Mathstud28
Quote:

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

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

sry for the confusion

Ok this is how I see it since you have dt

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

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

can you go from there?
• Apr 27th 2008, 08:48 AM
PaulRS
Quote:

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

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

sry for the confusion

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

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

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

;)
• Apr 27th 2008, 08:51 AM
tasukete
Quote:

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

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

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

;)

gosh that looks very complicated.. (Headbang)
• Apr 27th 2008, 08:53 AM
PaulRS
Quote:

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

That's not true, poor integral (Crying)

Note that $\displaystyle \left[ {\ln \left( {g + \tfrac{{b \cdot v}} {m}} \right)} \right]^\prime = \frac{{\tfrac{{b \cdot v'}} {m}}} {{g + \tfrac{{b \cdot v}} {m}}}$ (Wink)
• Apr 27th 2008, 09:08 AM
tasukete
Quote:

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

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

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

can you go from there?

im unsure what u just did there
• Apr 27th 2008, 09:29 AM
tasukete
can i ask for a confirmation. unsure whether this is correct direction to take..

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

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

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

$\displaystyle t = \frac{-v}{g} - \frac{m}{b}logv + c$
• Apr 27th 2008, 09:34 AM
Moo
Quote:

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

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

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

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

The inverse of $\displaystyle a+b$ is not $\displaystyle \frac 1a + \frac 1b$ ...
• Apr 27th 2008, 09:39 AM
tasukete
Quote:

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

oh my...stupid me
• Apr 27th 2008, 09:57 AM
tasukete
$\displaystyle \frac{dv}{dt} = -g - \frac{bv}{m}$

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

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

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

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