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Homework Help: Another First-Order ODE Question

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Going through a book on Numerical Methods, it states,

    [tex]y' = 4e^{0.8x}-0.5y[/tex]

    has the analytical solution,

    [tex]y= \dfrac{4}{1.3}(e^{0.8x}-e^{-0.5x})+2e^{-0.5x}[/tex]

    2. Relevant equations

    This is of the form,


    Should I use an Integrating Factor to solve the Linear ODE?, i.e. use an integrating factor [tex]\mu(x)=e^{\int{p(x)dx}}[/tex]




    ...and divide through by [tex]\mu(x)[/tex] ???

    3. The attempt at a solution



    and it is given: y(0) = 2

    Any pointers? Dick, your help would be much appreciated once again!

    Thanks and happy new year to all as well,

    Last edited: Jan 6, 2009
  2. jcsd
  3. Jan 6, 2009 #2


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    When computing your integrating factor what did you use as your p(t)? Your integrating factor is not correct.
  4. Jan 6, 2009 #3
    Note: with p(t)dt, consider it as p(x)dx. I used some old latex definition I had already typed for the 'method'. i.e., rather than dy/dt, this is dy,dx. dy/dt = f(t,y) -> dy/dx = f(x,y).

    *Definition updated in OP, w.r.t to (x).
    Last edited: Jan 6, 2009
  5. Jan 6, 2009 #4
    I used,

    [tex]\mu(x) = e^{\int{p(x)dx}}=e^{\int{\dfrac{1}{2}dx}} = e^{0.5x}[/tex]

    Well, isn't this,


    hence, [tex]p(x)=0.5[/tex] and [tex]q(x)=4e^{0.8x}[/tex] ???
    Last edited: Jan 6, 2009
  6. Jan 6, 2009 #5


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    [tex] e^{\int{\dfrac{1}{2}dx}} [/tex] Try doing this integral again. Where did the ln 2 come from?
  7. Jan 6, 2009 #6
    Oooops!!! thanks for spotting that blatant error. [tex]IF=e^{\int{p(x)dx}}=e^{0.5x}[/tex].
  8. Jan 6, 2009 #7
    This is what I am getting now,




    Where am I going wrong again?!?
    Last edited: Jan 7, 2009
  9. Jan 7, 2009 #8

    When I left the problem earlier as [tex]e^{0.5x}y=4\int{e^{0.8x}e^{0.5x}\,dx}+C[/tex], I thought I'd need to use integration by parts. But once I got it in the simpler form above (adding the powers) it just seems 'too simple' and the solution is missing a term.

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