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Math Help - differential equation help >.<

  1. #1
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    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.

    <br /> <br />
dv/dt = -g - bv/m<br /> <br />

    pls help me
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by tasukete View Post
    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.

    <br /> <br />
dv/dt = /int -g - bv/m<br /> <br />

    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?
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    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:

    <br />
\frac{dv}{dt}= -g -\frac{bv}{m}<br />

    sry for the confusion
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  4. #4
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    this is actually a physics question where b is the drag coefficient, g is gravity, m is mass, v is velocity
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  5. #5
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    im asked to find the velocity, v, in terms of t,b,g,m
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by tasukete View Post
    sry i typed it incorrectly. its:

    <br />
\frac{dv}{dt}= -g -\frac{bv}{m}<br />

    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?
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  7. #7
    Super Member PaulRS's Avatar
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    Quote Originally Posted by tasukete View Post
    sry i typed it incorrectly. its:

    <br />
\frac{dv}{dt}= -g -\frac{bv}{m}<br />

    sry for the confusion

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

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

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

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  8. #8
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    Quote Originally Posted by PaulRS View Post
    We have: <br />
\frac{{v'}}<br />
{{g + \tfrac{{b \cdot v}}<br />
{m}}} = - 1 \Rightarrow \int_0^t {\frac{{v'}}<br />
{{g + \tfrac{{b \cdot v}}<br />
{m}}}dt} = - t<br />

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

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

    gosh that looks very complicated..
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  9. #9
    Super Member PaulRS's Avatar
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    Quote Originally Posted by tasukete View Post
    gosh that looks very complicated..
    That's not true, poor integral

    Note that <br />
\left[ {\ln \left( {g + \tfrac{{b \cdot v}}<br />
{m}} \right)} \right]^\prime   = \frac{{\tfrac{{b \cdot v'}}<br />
{m}}}<br />
{{g + \tfrac{{b \cdot v}}<br />
{m}}}<br />
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    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
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  11. #11
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    can i ask for a confirmation. unsure whether this is correct direction to take..

    <br />
\frac{dv}{dt} = -g - \frac{bv}{m}<br />


    <br />
\frac{dt}{dv} = \frac{-1}{g} - \frac{m}{bv}<br />


    <br />
t = \int \frac{-1}{g} - \frac{m}{bv}dv<br />


    <br />
t = \frac{-v}{g} - \frac{m}{b}logv + c<br />
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  12. #12
    Moo
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    Quote Originally Posted by tasukete View Post
    can i ask for a confirmation. unsure whether this is correct direction to take..

    <br />
\frac{dv}{dt} = -g - \frac{bv}{m}<br />


    <br />
\frac{dt}{dv} = \frac{-1}{g} - \frac{m}{bv}<br />


    <br />
t = \int \frac{-1}{g} - \frac{m}{bv}dv<br />


    The inverse of a+b is not \frac 1a + \frac 1b ...
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  13. #13
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    Quote Originally Posted by Moo View Post
    The inverse of a+b is not \frac 1a + \frac 1b ...

    oh my...stupid me
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  14. #14
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    <br />
\frac{dv}{dt} = -g - \frac{bv}{m}<br />


    <br />
\frac{dv}{dt} = \frac{-gm - bv}{m}<br />


    <br />
\frac{dt}{dv} = \frac{m}{-gm - bv}<br />


    <br />
t = \int \frac{m}{-gm - bv}dv<br />


    <br />
t = -m log(gm + bv) + c<br />
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