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Math Help - Itegration

  1. #1
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    Itegration

    Integrate the following equation with resprect to x twice:

    f'(x) + x f''(x) = 0
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  2. #2
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    Hello, feage7!

    A slightly different approach . . .


    Integrate with resprect to x twice: . f'(x) + x f''(x) \:= \:0

    Note that we have: . \frac{d}{dx}\left[x\!\cdot\!f'(x)\right] \:=\:0

    Integrate: . x\!\cdot\!f'(x)\:=\:C_1


    Integrate again: . \int x\!\cdot\!f'(x)\,dx\:=\:\int C_1\,dx

    By parts:
    . . \begin{array}{cc}u = x & dv = f'(x)\,dx \\ du = dx & v = f(x)\end{array}

    And we have: . x\!\cdot\!f(x) - \int f(x)\,dx\:=\:C_1x + C_2

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  3. #3
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    ok ive tried using that and it didnt help much in terms of what i needed to do so ill try n show the full question, it is a dimensional analysis question but its the integration part im having difficulty on;

    you have the following variables: t, R, p, M, e, y, u.

    using bukcinghams pi theorem u get the equation

    u/ ((t/p)^1/2)=f(Rp/M x (t/p)^1/2 , y/R , e/R)

    then by adapting a new constraint another equation is formed basically the same as above only e/R is removed and y/R is now known as n and Rp/M x (t/p)^1/2 is known as E :

    u/ ((t/p)^1/2)=f1(En)

    then another equation is derived:

    f3(E)=f1(En)+f2(n)

    differentiate once with respect to E, and then differentiate that answer with respect to n to get the equation

    f1'(En) + En f1"(En)=0

    let x = En

    and integrate twice with respect to x to obtain the equation:

    u/((t/p)^1/2)= A ln{y/m x p (t/p)^1/2 } + B

    dnt know if ive gave you enough information there
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