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Differential equuation with boundary conditions

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data
    d^2T/dx^2+S/K=0 Boundary Conditions T=Tsub1 @ x=0
    and T=Tsub2 @ x=L


    2. Relevant equations



    3. The attempt at a solution

    d^2T/dx^2 = -(S/K) <--- intergrate to get
    dT=-(S/K)dx+ C1 <--- intergrate to get
    T=(-S/K)x+c1+c2
    apply both boundary conditions to get
    Tsub1=c1+c2

    Not sure if i doing it right and if i am i dont know how to get c1 and c2
    Thanks for your help
    Tsub2=-(S/K)*L +c1+c2
     
  2. jcsd
  3. Aug 24, 2010 #2

    gabbagabbahey

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    You mean "integrate", right :wink: And you aren't integrating to get to here, you are simply rearranging your DE algebraically.

    No,

    [tex]\frac{d^2T}{dx^2}=-\frac{S}{K} \implies \frac{dT}{dx}=\int -\frac{S}{K}dx= -\frac{S}{K}x+C_1[/tex]

    You will need to integrate once more to get [itex]T[/itex]
     
  4. Aug 24, 2010 #3
    I meant integrate that equation to get to the next line.

    And after I integrate I get: T=-(S/K)* (x^2)/2+C1+C2

    Now I just plug in the boudary conditions and solve the system of equations for c1 and c2 correct?
     
  5. Aug 24, 2010 #4

    gabbagabbahey

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    [tex]\int(-\frac{S}{K}x+C_1)dx\neq -\frac{S}{K}x^2+C_1 +C_2[/tex]

    You're missing something.
     
  6. Aug 24, 2010 #5
    =-S/K* x^2/2+C1x+C2

    Also how do I make my equations appear like yours?
     
  7. Aug 24, 2010 #6

    gabbagabbahey

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    Good, now use your boundary conditions to find C1 and C2

    Click on the link in my sig :wink:
     
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