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Solve differential equation

  1. Nov 2, 2011 #1

    syj

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    1. The problem statement, all variables and given/known data

    [itex]\frac{dr}{dt}=r(u-r)[/itex]

    2. Relevant equations

    i rewrote this as [itex]\frac{dr}{dt}-ur=-r^2[/itex]

    i think this is like bernoulli's equation



    3. The attempt at a solution
    so i let [itex]w=\frac{1}{r}[/itex] so that [itex]r=\frac{1}{w}[/itex]
    this gives me

    [itex]\frac{dr}{dt}=-1w^{-2}\frac{dw}{dt}[/itex]

    so now i have
    [itex]-w^{-2}\frac{dw}{dt}-uw^{-1}=-w^{-2}[/itex]

    if i divide by [itex]-w^{-2}[/itex]

    then i get a linear differential equation

    [itex]\frac{dw}{dt}+uw=1[/itex]

    the integrating factor is [itex]e^{ut}[/itex]

    so i get [itex]w=\frac{1}{u}+w_0e^{-ut}[/itex]

    from my definition of [itex]w[/itex] i have [itex]w_0=\frac{1}{r_0}[/itex] and [itex]w=\frac{1}{r}[/itex]

    so i get [itex]r=\frac{ur_0}{r_0+ue^{-ut}}[/itex]

    is this right?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 2, 2011 #2

    lanedance

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    Homework Helper

    check by differentiating and comparing to the original DE

    note this equation is separable, so you could do it by direct integration
     
  4. Nov 2, 2011 #3

    syj

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    please explain how i would solve by separation.
    thanks
     
  5. Nov 2, 2011 #4

    lanedance

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    Homework Helper

    [tex]
    \frac{dr}{dt}=r(u-r)
    [/tex]

    gives
    [tex]
    \int \frac{dr}{r(u-r)}=\int dt
    [/tex]

    though even if you can solve the integral (partial fractions may be a good start) it may be difficult to get into an analytic function for r(t)
     
  6. Nov 2, 2011 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    [tex]\frac{dr}{dt}= r(u-r)[/tex]
    with u constant.

    [tex]\frac{dr}{r(u-r)}= dt[/tex]

    Integrate the left side with respect to r and the right side with respect to t.

    ( I really hate it when lanedance gets in one or two minutes ahead of me but to have him first with exactly the same timestand- arrrgh!)
     
  7. Nov 3, 2011 #6

    syj

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    ok so i get
    [itex]\frac{ln(r)-ln(r-u)}{u}=t[/itex]

    which i then manipulate (or try to at least ) to get:
    [itex]
    ln\frac{r}{r-u}=ut
    [/itex]

    if this is correct then i think i can do:
    [itex]
    \frac{r}{r-u}=e^{ut}
    [/itex]

    i know something is wrong cos i dont have [itex] r_0 [/itex]

    im working through an example in jordan and smith (nonlinear ordinary differential equations)
    so i know what the answer should be, im trying to get to it tho.
    :(
    please please correc this.
     
    Last edited: Nov 3, 2011
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