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Simple-looking 1st order DE, solution anyone?

  1. Apr 15, 2013 #1
    I have a differential equation
    [tex]
    \frac{dx}{dt} = \frac{a-x}{b-t} - cx
    [/tex]
    but no idea if there is an analytical solution :(
    I tried the integrating factor method, but I run into a e^(-x)/x^2 kind of integral, which I think doesn't have an analytical expression.. Help, please?
     
  2. jcsd
  3. Apr 15, 2013 #2

    ShayanJ

    User Avatar
    Gold Member

    With some manipulation,your equation can get the form below:
    [itex]
    -(t-b)(\dot{x}+cx)+x=a
    [/itex]
    Now take a solution like:
    [itex] x=\sum_0^{\infty}a_n(t-b)^{n+r}[/itex]
    We have:
    [itex]
    \dot{x}=\sum_0^{\infty}a_n (n+r) (t-b)^{n+r-1}
    [/itex]
    If you substitute the x and [itex] \dot{x} [/itex] in the homogenous DE,you will find:
    [itex]
    -\sum_0^{\infty}a_n(n+r)(t-b)^{n+r}+\sum_0^{\infty}a_n(t-b)^{n+r}-\sum_1^{\infty}ca_{n-1}(t-b)^{n+r}=0
    [/itex]
    Where I have changed the summation index in the third term from n to n-1 to make the powers of (t-b) the same in all summations.
    Now to make the summations start from a common n,I get the n=0 terms out of the first two summations:
    [itex]
    -a_0r(t-b)^r+a_0(t-b)^r=0
    [/itex]
    Which gives [itex] r=1 [/itex]
    Then you can take all the terms in one summation and equate the coefficient of [itex] (t-b)^{n+r}[/itex] to zero to get a recursion relation which determines the series solution.
    Then you should add a/(c+1) to get the solution to the in-homogenous equation.
     
    Last edited: Apr 15, 2013
  4. Apr 15, 2013 #3
    so you mean to say that there is no solution in terms of elementary functions?
     
  5. Apr 15, 2013 #4

    ShayanJ

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    Gold Member

    The series you find may come out to be the power series corresponding to an elementary function.But it only may.
     
  6. Apr 15, 2013 #5
    argh :(
    better integrate numerically then...
     
  7. Apr 15, 2013 #6

    ShayanJ

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    Gold Member

    I just used www.wolframalpha.com to solve your equation and this is the result:
    http://www3.wolframalpha.com/Calculate/MSP/MSP1811gi6c448044hadb10000242868gf692egc43?MSPStoreType=image/gif&s=12&w=389.&h=22. [Broken]
    With [itex]x \rightarrow t[/itex] and [itex]y \rightarrow x[/itex]
    And Ei(x) is the exponential integral which you can find its values in some tables of functions.
     
    Last edited by a moderator: May 6, 2017
  8. Apr 15, 2013 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    That is a linear differential equation.
    [tex]\frac{dx}{dt}+ (\frac{1}{b- t}+ c)x= \frac{a}{b- t}[/tex]
    An "integrating factor" is a function u(t) such that
    [tex]\frac{dux}{dt}= u\frac{dx}{dt}+ u(\frac{1}{b- t}+ c)x[/tex]
    Since
    [tex]\frac{dux}{dt}= u\frac{dx}{dt}+ \frac{du}{dt}x[/tex]
    that means we must have
    [tex]\frac{du}{dt}= \frac{1}{b- t}+ c[/tex]
    [tex]du= \left(\frac{1}{b- t}+ c\right)dt[/tex]

    u(t)= -ln|b-t|+ ct

    That is, the original differential equation is
    [tex]\frac{d}{dt}\left(x(t)(-ln|b- t|+ c\right)= (-ln|b- t|+ ct)\frac{a}{b- t}[/tex]
    so that
    [tex]x(t)(-ln|b- t|+ c)= \int (-ln|b-t|+ ct)\frac{a}{b- t} dt[/tex]

    Let u= b- t so that du= dt and the right side becomes
    [tex]b\int \frac{ln|u|}{u}du- ac\int \frac{b- u}{u}du[/tex]
    In the first integral let v= ln|u| so that dv= du/u and that integral becomes
    [tex]\int v dv= (1/2)v^2= (1/2)(ln|u|)^2= (1/2)(ln|b- t|)^2[/tex]
    In the second integral (b-u)/u= (b/u)- 1 and its integral bln|u|- u= b ln|b- t|- |b- t|.
    putting those together
    [tex]x(t)(-ln|b- t|+ c)= (1/2)(ln|b-t|)^2+ b ln|b- t|- |b- t|)[/tex]

    Finally,
    [tex]x(t)= \frac{ln|b-t|)^2+ b ln|b- t|- |b- t|}{2(-ln|b- t|+ c)}[/tex]
     
    Last edited: Apr 15, 2013
  9. Apr 15, 2013 #8
    This can't be right. If you set c=0, the solution is simply x= at/b, and your equation doesn't reduce to that. Also, if b -> infinity, the solution tends to x = Ae(-ct), which I see no trace of in your calculations.
     
    Last edited: Apr 15, 2013
  10. Apr 15, 2013 #9

    ShayanJ

    User Avatar
    Gold Member

    Here's the problem:
    We should have:
    [itex]
    \frac{du}{dt}=u(\frac{1}{b-t}+c)
    [/itex]
    Which gives an integrating factor of:
    [itex]
    u=\frac{e^{ct}}{b-t}
    [/itex]
    And we will have:
    [itex]
    \frac{xe^{ct}}{b-t}=\int \frac{ae^{ct}}{(b-t)^2}dt
    [/itex]
    And this gives the answer that I tried to present in the post #6 but it seems that its image is deleted from the server of the wolframalpha.
    The answer is:
    [itex]
    -{{\rm e}^{-ct}}a \left( -{{\rm e}^{ct}}+{{\rm e}^{cb}}{\it Ei}
    \left( 1,cb-ct \right) cb-{{\rm e}^{cb}}{\it Ei} \left( 1,cb-ct
    \right) ct \right)
    [/itex]
    Where:
    [itex]
    Ei(a,z)=\int_1^{\infty} e^{-tz}t^{-a}dt
    [/itex]
    is the exponential integral and you can find its values in some mathematical tables.
     
    Last edited: Apr 15, 2013
  11. Apr 23, 2013 #10
    Just trying :|

    Isn't it a linear DE? I'm not really good at these kinds of things, but here's my try:

    I'm not sure, I don't really go for analytic solutions :|


    EDIT: I looked at how wolframalpha would integrate the right-hand integral and it generated an Ei(-c(t-b)) o-o
     

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    Last edited: Apr 23, 2013
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