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2nd Order, homogeneous Differential Equation

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve d2θ/dη2 + 2η(dθ/dη) = 0, to obtain θ as a function of η,
    where θ=(T-T0)/(Ts-T0)

    EDIT: I should add that this is a multi-part problem, and η is given as η=Cxtm. We had to use that to derive the equation in question above.. So I dont know if this is supposed to be solved as a non-constant coefficient method or not... My method below solved it assuming η was a constant.. I can supply the whole problem as an attachment if necessary.

    2. Relevant equations
    ay"+by'+cy=0
    ar2+br+c=0
    If the roots are real and different, solution is: y=aer1x+ber2x

    3. The attempt at a solution
    I would assume this can just be:
    θ"+2ηθ'=0
    which turns to:
    r2+2ηr=0

    But when using the quadratic equation to get roots, I get r1=-2η, and r2=0

    Plug this into the solution form and get θ=ae-2ηx

    Not sure if this is right. Can someone confirm, or tell me what I'm doing wrong? Thanks!
     
    Last edited: Apr 20, 2015
  2. jcsd
  3. Apr 20, 2015 #2

    Orodruin

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    If your function is a function of ##\eta##, you clearly cannot make the assumption that ##\eta## is a constant.

    The differential equation you have for ##f(\eta) = d\theta/d\eta## is separable.
     
  4. Apr 20, 2015 #3
    Okay, so all I know about separable equations is that you have to get like terms on the same side of the equation.
    If I rearranged the original equation, it would give me:
    d2θ/dθ = -2η(d2η/dη)

    But I have no clue how to solve that.
     
  5. Apr 20, 2015 #4

    Orodruin

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    No, the equation for ##f = d\theta/d\eta## is separable. Solve that first and then integrate it to find ##\theta##.
     
  6. Apr 20, 2015 #5
    I dont have an equation for dθ/dη. Are you referring to the equation I have for θ in terms of T ( θ=(T-T0)/(Ts-T0) )? Because I'm not sure how to differentiate that with respect to η to obtain the dθ/dη equation...
    I apologize for my very rusty differential equations abilities, and I appreciate you trying to help.
     
  7. Apr 20, 2015 #6

    Orodruin

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    Make the substitution ##f = d\theta/d\eta##. This gives you a differenial equation for ##f## that you have to solve.
     
  8. Apr 20, 2015 #7
    Okay I think I made progress. (crossing fingers)..
    Using the substitution f=dθ/dη, that means df/dη=d2θ/dη2.
    Plugging these into the equation d2θ/dη2 + 2η(dθ/dη) = 0 yields:
    ⇒df/dη + 2ηf =0
    ⇒df/dη = -2ηf
    ⇒(1/f)df = -2η(dη)

    Integrate both sides:
    ⇒∫(1/f)df = ∫-2η(dη)
    ⇒ln(f) = -η2

    Take base 'e' of both sides to get:
    f=e2

    recalling that f=dθ/dη,
    ⇒dθ/dη = e2

    (This is where I may be wrong)... Rearrange and integrate again??
    dθ = e2
    ⇒∫dθ =∫e2

    θ = ½*√π*erf(η)+C (I used Wolfram Integral Calculator for this)
     
  9. Apr 20, 2015 #8

    Orodruin

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    Yes, apart from that you missed an integration constant in the first integration.
     
  10. Apr 20, 2015 #9
    Okay yes, I missed the constant. So if I put that in there, the first integral would become:
    f=e2+C

    ⇒dθ/dη = e2+C
    ⇒dθ = [e2+C]dη

    Integrate:
    ∫dθ =∫[e2+C]dη
    ⇒θ = ½*√π*erf(η)+η+C

    I think my answer would change to include an additional η, but does that error function even make sense? I feel like that doesnt make solving the temperature profile any easier, if that were the objective in a real-life application..
     
  11. Apr 20, 2015 #10

    Orodruin

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    Not really. The constant shows up before you exponentiate the result. It is therefore a multiplicative constant rather than additive.
     
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