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Math Help - Reducing order of DFQ and solving

  1. #1
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    Reducing order of DFQ and solving

    Initially I have a function like this:

    f''' + 1 - (f')^2 = 0 ... and f' (is f prime)

    To solve this I am told to multiply by an integrating factor f'' and then integrate. So I get:

    1/2(f'')^2 + f' - 1/3(f')^3 = C

    And then the problem says to make a substitution that G = f' so I get:
    1/2(G')^2 + G - 1/3(G)^3 = C ...which is a a 1st order ODE but I don't know how to solve that. Can anyone help me?

    Thanks
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  2. #2
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    It looks already separable to me. Just treat f' as a variable so that:

    f''+1-(f')^2=0\Rightarrow d(f')+1-(f')^2=0

    Separating I get:

    \frac{d(f')}{(f')^2-1}=dx

    Integrate, isolate f' then integrate again.
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  3. #3
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    Quote Originally Posted by shawsend View Post
    It looks already separable to me. Just treat f' as a variable so that:

    f''+1-(f')^2=0\Rightarrow d(f')+1-(f')^2=0

    Separating I get:

    \frac{d(f')}{(f')^2-1}=dx

    Integrate, isolate f' then integrate again.
    Sorry the initial equation is :

    f'''+1-(f')^2=0 It was probably hard to read before

    The problem says to first multiply by a integrating factor  f''

    So I get to:

     1/2 (f'')^2 + f' -1/3(f')^2 = C

    I am then told to make the substitution  G(n) = f'(n) solve this to show that

     f' = 3*tanh^2[n/root2 + 1.146] -2

    f in all of this is a function of n, and the boundary conditions are:

    f(0)= f'(0) =0 , and f'(n) approaches 1 as n approaches infinity

    Please someone help
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