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Homework Help: 2nd order ODE

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] \frac { d^2 \theta }{d x'^2 } = -y *exp(\theta) [/tex] eq. 1

    mayb be integrated to yield

    [tex] exp(\theta) = \frac {a}{cosh^2(b \frac{+}{-} \sqrt \frac{a*y}{2} * x')} [/tex]

    [tex] \theta = f(y,x') [/tex]

    2. Relevant equations

    3. The attempt at a solution

    the exponent is throwing me off, but i probably have to use the following properties

    [tex] \frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(/theta) = 0 [/tex]

    [tex] g = [1/2 (e^k^x + e^-^k^x)]^2 = cosh^2(kx) [/tex]

    the problem is the exp(theta), do I have to replace the variable with something like

    g = ln(theta)

    then find

    [tex] \frac { d^2 g }{d x'^2 } [/tex]

    to get

    [tex] \frac { d^2 g }{d x'^2 } + \sqrt{y}*g = 0 [/tex]

    then apply

    m^2 + k^2 = 0?

  2. jcsd
  3. Nov 23, 2009 #2


    Staff: Mentor

    I'm confused. In the other thread you started, you have
    [tex] \frac { \partial^2 \theta }{\partial x'^2 } = -y *exp(\theta) [/tex]
    and in this one you have it as a 2nd-order ODE.
    Since theta is a function of y and x', you really need partial derivatives, not ordinary derivatives. If theta were a function of a single variable, say x', then you could talk about d(theta)/dx' and d^2(theta)/dx'^2.

    Having said that, how do you go from equation 1 to this equation? IOW, how did y in the first equation become sqrt(y) in this equation?
    [tex] \frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(\theta) = 0 [/tex]
  4. Nov 24, 2009 #3
    sorry, at first it was a partial derivative but then I realized that theta is only dependent on the spatial variable x', like you said. Since this is a steady state problem (conditions for the problem approaching a steady state are given by the rate of reaction, it reduces to a ODE


    I found more info on how to solve the equation

    [tex] x' = \int \frac{d \theta}{\sqrt{-2*\int -y*exp(\theta) d \theta}} [/tex]

    then i get

    [tex] x' = \int \frac {d\theta}{\sqrt{2*y*exp(\theta) + 2*y*c}} [/tex]

    and now im trying to figure out the above integral.
    Last edited: Nov 24, 2009
  5. Nov 24, 2009 #4


    Staff: Mentor

    Factor 2y out of the terms in the radical and bring them out of the radical and out of the integral so that you have exp(theta) + c in the radical. I would then see if a substitution would work.
  6. Nov 24, 2009 #5
    ok so following ur suggestion i did

    [tex] x' = \frac {1}{\sqrt{2y}} * \int \frac {d\theta}{\sqrt{\int exp(\theta)d\theta}} [/tex]

    [tex] x' = \frac {1}{\sqrt{2y}} * \int \frac {d\theta}{\sqrt{ exp(\theta)+c}} [/tex]

    then i use the identity for int 1/sqrt(theta^2 + a^2) that is

    [tex] x' = \frac {1}{\sqrt{2y}} * ln | exp(\theta/2) + \sqrt{exp(\theta) + c} | + a [/tex]

    but when i try to solve for theta, i dont know how to reduce the exp to fit the form of the solution the book gives for exp(theta)
  7. Nov 24, 2009 #6


    Staff: Mentor

    Not sure if this will help, but the integral formula you used can be expressed another way.
    [tex]\int \frac{dx}{\sqrt{x^2 + a^2}}~=~sinh^{-1}(x/a) + C[/tex]

    One thing I've been bothered by is treating y as if it were a constant in the integration with respect to theta. You have theta = f(y, x'), and in the other thread you said that y = x/x'.
  8. Nov 24, 2009 #7
    hey mike, thanks so much for your hlep. yes that identity definitely helps.

    please try to ignore the other post

    y here is known as the frank-kamenetskii parameter

    [tex] y = \frac {E}{RT^2_a} \frac {Q}{h} *r^2*z*exp(-E/RT_a) [/tex]

    now, the r^2 in the y comes from x' = x/r

    this is the substitution to drop the units of x, where r can be the half width or radius of a cylinder.

    specifically, by the chain rule

    [tex] \Delta x' = r^2 \Delta [/tex]

    Quoting the book, In the simplest problems, the boundary condition is \theta = 0 at the surface, and the critical condition reduces to [tex] y = const = y_c_r_i_t [/tex] since neither in the equation nor in the boundary condions are there any parameters other than y.

    So i think its ok to factor out y, no?
    Last edited: Nov 24, 2009
  9. Nov 24, 2009 #8


    Staff: Mentor

    That's Mark...
    You have [tex] \Delta x' = r^2 \Delta [/tex]
    Don't you mean this?
    [tex] \Delta x' = \Delta r^2[/tex]
    Based on what you said about y, sounds like you can move it out of the integral.
  10. Nov 24, 2009 #9
    sorry, mark :)

    I still don't get the same answer as in the book

    I did

    [tex] x' = \frac {1}{\sqrt{2y}} * sech^-^1 (\frac {exp(\frac {\theta}{2})}{\sqrt{c}}) + a [/tex]

    when I solve for [tex] exp(\theta) [/tex] i get

    [tex] exp(\theta) = c* sech^2(\sqrt{2y}*x' + a) [/tex]


    [tex] exp(\theta) = \frac {2c}{cosh^2 (\sqrt {2y}*x' + a)} [/tex]

    the only problem now are the constants, the book has

    [tex] exp(\theta) = \frac {a}{cosh^2 ( \sqrt {\frac {ay}{2}}*x' \frac {+}{-} b)} [/tex]

    im confused why they have +/- and why it is sqrt(ay)/2
    Last edited: Nov 24, 2009
  11. Nov 24, 2009 #10


    Staff: Mentor

    1/cosh^2(x) = sech^2(x)
  12. Nov 24, 2009 #11
    [tex] exp(\theta) = \frac {2c}{cosh^2 (\sqrt {2y}*x' + a)} [/tex]

    the only problem now are the constants, the book has

    [tex] exp(\theta) = \frac {a}{cosh^2 ( \sqrt {\frac {ay}{2}}*x' \frac {+}{-} b)} [/tex]

    im confused why they have +/- and why it is sqrt(ay/2)
  13. Nov 24, 2009 #12


    Staff: Mentor

    Got me, so that looks like something you'll have to puzzle out...
  14. Nov 24, 2009 #13

    well, thanks again for getting me this close. great help :)
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