1. 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

2. 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?

3. 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

4. this is actually a physics question where b is the drag coefficient, g is gravity, m is mass, v is velocity

5. im asked to find the velocity, v, in terms of t,b,g,m

6. 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?

7. 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}}}}
$

8. 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..

9. Originally Posted by tasukete
gosh that looks very complicated..
That's not true, poor integral

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}}}
$

10. 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

11. 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
$

12. 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 $\frac 1a + \frac 1b$ ...

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

oh my...stupid me

14. $
\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
$