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1st Order Linear ODE integrating factors

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi, I submitted this question on here the other day a user suggested some topics which might help so I have went away and tried this and this is what I have came up with. I just want to know what I have so far is right also I need help with integrating the rhs of the equation.

    The question is
    dy/dx + 0.8y = 0.6 e-1.4x y(0)=1

    I have to solve the ODE to the given condition using exact methods and evaluate the solution y for x = 0.0 to x=0.5 in steps of 0.05.


    2. Relevant equations

    dy/dx + P(x)y = Q(x)



    3. The attempt at a solution

    so I my working so far is

    dy/dx + 0.8y = 0.6 e^-1.4x

    P(x) = 0.8

    IF = e ^ ∫P(x) dx
    IF = e ^∫0.8 dx
    = e ^ 0.8x

    e0.8x* dy/dx + e0.8xy = e0.8x*0.6e-1.4x

    Therefore e0.8x = ∫ e0.8x*0.6e-1.4x

    This is where I get stuck. What I have done so far is this correct and what should I do to progress from this point? I don't understand how to do the integration on the rhs.
     
  2. jcsd
  3. Nov 6, 2011 #2

    LCKurtz

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    What happened to the y in that last step? Also you can use rules of exponents to simplify the integrand.
     
  4. Nov 6, 2011 #3
    This is what I had
    e0.8x = ∫ e0.8x*0.6e-1.4x

    So should it be this instead where the multiplying exponents just add together?

    e0.8xy = ∫0.6e-2.2x

    What would happen now?
    Would the rhs integrated become 0.6 ln(-2.2x) ?
     
  5. Nov 6, 2011 #4

    LCKurtz

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    Yes, but did you add them together correctly?
    No. If you are taking differential equations, you should by now know your antiderivatives. Look up the antiderivative of eax in your calculus book.
     
  6. Nov 6, 2011 #5
    Ok I see that one of the exponents was -1.4 so adding that 0.8 will give us 0.6.
    I have checked out my notes I just have a basic list of integrals and can't seem to find anything similar online for eax so would it be similar to eax+b where I would just miss out the plus b part? So it would just 1/0.6 e 0.6 + C? Also what would happen to the original 0.6 before the e on the rhs of the equation.
     
  7. Nov 6, 2011 #6

    LCKurtz

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    I'm curious what level of mathematics you are taking. Are you reading ahead of the courses you are taking? You still haven't successfully added -1.4 and 0.8. And you simplify whatever you get after you solve for y.
     
  8. Nov 6, 2011 #7
    My level of maths is not too good I have to admit. I was on an easier degree course last year which had a background in the built environment it still had some maths and physics elements but wasn't as heavy on integration and differentiation. The course transferred over to a harder engineering degree and I'm totally lost on this maths module, but i'm trying to pick my way through it just to get a basic pass on the module as its the only module on my course that I have difficulty with.

    So thats -0.6 ?

    If not I give up ha
    but thanks for your help anyway!
     
  9. Nov 6, 2011 #8

    LCKurtz

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    Yes. -1.4+.8 = -0.6. I would guess that you are headed for a lot of frustration and possible failure if you are taking an engineering degree without adequate mathematics training. Good luck with that.
     
  10. Nov 6, 2011 #9
    You're right i'm pulling my hair out with it already, i'll get there in the end though! Just one last thing though. How would the anti derivative e^ax work? I'll be harassing my lecturers tomorrow haha.
     
  11. Nov 6, 2011 #10

    LCKurtz

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    The derivative of eax= aeax. So the antiderivative of eax is (1/a)eax+C, assuming of course, that a isn't 0.
     
  12. Nov 6, 2011 #11
    Thanks for your time. I appreciate it.
     
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