# Thread: differential equation help >.<

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.

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

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.

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

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?

3. 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

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:

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

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?

7. 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

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

8. 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..

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

Note that $\displaystyle \left[ {\ln \left( {g + \tfrac{{b \cdot v}} {m}} \right)} \right]^\prime = \frac{{\tfrac{{b \cdot v'}} {m}}} {{g + \tfrac{{b \cdot v}} {m}}}$

10. 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

11. can i ask for a confirmation. unsure whether this is correct direction to take..

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

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

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

$\displaystyle t = \frac{-v}{g} - \frac{m}{b}logv + c$

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

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

oh my...stupid me

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

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

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

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

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