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Determine the function so that the integral holds

  1. Mar 9, 2008 #1


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    1. The problem statement, all variables and given/known data

    Determine all functions [tex]H:(0,\infty)\to\mathbb{R}[/tex]
    so that [tex]H(x)=\frac{1}{x}\int_{2}^{x}t(3-2H'(t))\ dt,\ \ x>0[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I tried to integrate by parts to change the integral into to the form of a differential equation, or at least an equation that could be differentiated to yield a differential equation. However; I always end up with a 1/x term in from of my integral thus making it impossible to get only H'(x) and H(x). Basically, I can't eliminate the integral of H(x) from my equation
  2. jcsd
  3. Mar 9, 2008 #2
    try to differentiate both parts with respect to x, and see if you can get anything. or just try to integrate the right-handed side and see if you can come up with sth.
  4. Mar 9, 2008 #3


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    I had to multiply the equation with x then the differentiation works out fine
  5. Mar 9, 2008 #4
    Yeah, after that you should have gotten a differential equation, i think it should be a linear one, which is solved using an integrating factor. So all solutions of that diff. eq are indeed all the functions H(t)=(3/4) x+Cx^-3, if i can remember it well, because i did it yesterday or sth after you posted, and i am not sure if this is the exact answer i got.
    But as a check to your answer take the derivative of this function, and plug it in the integral, integrate it and see if you get the same function H(t).
  6. Mar 10, 2008 #5


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    yeah its a linear one and the result I got is H=(3/4)x + Cx^(-1/3)
  7. Mar 10, 2008 #6
    are u sure u didn't do any mistakes on your way!!!??i think the answer should read

    [tex] H(x)=\frac{3}{4}x+Cx^{-3}[/tex] and not

    [tex] H(x)=\frac{3}{4}x+Cx^{\frac{-1}{3}[/tex]

    as a means of checking your answer, follow the instructions i gave u in post # 4

    These kind of problems are quite interesting, this one too!! .....lol.....
    Last edited: Mar 10, 2008
  8. Mar 10, 2008 #7


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    I tried in maple
    [tex]dsolve(H(x)+x*(diff(H(x), x)) = x*(3-2*(diff(H(x), x))));[/tex]

    and got
    [tex] H(x) = \frac{3}{4}x + \frac{_C1}{x^{1/3}} [/tex]

    so I guess the 1/3 is correct.
    I did the exercise in class today the one thing I forgot was that because I differentiate I relax the constraints and have to require
    H(2) = 0
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