Integrating the Function: $\frac{dv}{dt}=g-\frac{c}{m}v$

[bortonj88]

Hi everyone

I am looking at integrating this function,

\frac{dv}{dt}=g-\frac{c}{m}v

So far i have rearranged the function to get,

(\int1/g - \int1/(cv/m))* dv= \intdt

Thanks

JB

 

[LeonhardEuler]

Are you saying:
\frac{dv}{dt}=g-\frac{c_V}{m}
and this rearranges to
\int\frac{1}{g}dv -\int\frac{1}{\frac{c_V}{m}}dv=\int dt
If so, there is a mistake because you can't split apart a denominator like that.

 

[Mark44]

Hi everyone

I am looking at integrating this function,

\frac{dv}{dt}=g-\frac{c}{m}v

So far i have rearranged the function to get,

(\int1/g - \int1/(cv/m))* dv= \intdt

Thanks

JB

There are at least a couple of ways of solving this differential equation, one of which is separation.
\frac{dv}{dt} = g -\frac{c}{m}v

\Rightarrow \frac{dv}{g -\frac{c}{m}v} = dt

\Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt

Tip: Use only one pair of tex tags per line rather than sprinkling them in the line.

 

[bortonj88]

yeah, sorry, i made a mistake with that.

It was meant to read;

\int(1/(g-(c/m)v)) dv = \intdt

Thanks
JB

 

[bortonj88]

<br /> \Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt<br />

Yeah i had this, i then integrated to get
<br /> \Rightarrow ln{gm -cv} = \frac{1}{m}t<br />

 

[bortonj88]

sorry, meant to read

<br /> <br /> \Rightarrow ln({gm -cv}) = \frac{1}{m}t<br /> <br />

 

[Mark44]

<br /> \Rightarrow \frac{dv}{gm -cv} = \frac{1}{m}dt<br />

Yeah i had this, i then integrated to get
<br /> \Rightarrow ln{gm -cv} = \frac{1}{m}t<br />

What was the substitution you used on the left side? Also, where is the constant of integration on the right side?

ln(gm -cv) = \frac{1}{m}t

 

[bortonj88]

thats the problem, i couldn't figure out how to do the integral on the left hand side, so I used quickmath.

If you have a suggestion please let me know, as this isn't homework, my housemate gave it to me to try and solve and now its really bugging me!

As for the constant, inital conditions are v=0 when t=0, and therefore the constant is not used, or so i thought

 

[Mark44]

The substitution is u = gm - cv, so du = -cdv. You need a factor of -c on the left, which you can get by multiplying the numerator by -c and also multiplying by -1/c.

Here's the left side.

\int \frac{dv}{gm -cv} = \frac{-1}{c}\int \frac{-cdv}{gm -cv}
= \frac{-1}{c}\int \frac{du}{u}

Can you finish it. Don't forget to undo the substitution.

Regarding the constant, it's safer to put it in, and then if it's zero you can remove it.