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ODE, bernoulli equation -> leads to crazy integral !

  1. Sep 7, 2011 #1
    ODE, bernoulli equation --> leads to crazy integral !

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

    An Initial Value Problem, ODE (Bernoulli equation)

    ODE: [x^2]*y' + 2*[x^3]*y = [y^2]*(1+2*[x^2])
    IV: y(1) = 1/2

    2. Relevant equations

    general form of Bernoulli's equation:
    y' + a(x)y = b(x)*[y^n]

    First order, linear ODE form:
    y' + a(x)y = b(x)

    3. The attempt at a solution

    I begin by simplifying the Bernoulli-type equation into a linear form, at which point I end up with an integral that I DO NOT KNOW HOW TO SOLVE.

    9k85f6.jpg

    I have either really messed up somewhere along the way to that Integral, or not and I just dont' know how to solve it.

    Thanks for any help.
     
  2. jcsd
  3. Sep 7, 2011 #2

    tiny-tim

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    Hi Jonnyb42! :smile:

    Looks ok to me …

    but you can't solve that integral unless you're allowed to use the "error function" erf(x). :redface:
     
  4. Sep 7, 2011 #3
    Re: ODE, bernoulli equation --> leads to crazy integral !

    No further ideas here, sorry :-\
     
  5. Sep 7, 2011 #4

    HallsofIvy

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    Re: ODE, bernoulli equation --> leads to crazy integral !

    Are you required to actually integrate it? I remember in my first D.E. class, I had one problem in the text that I could reduce to an integral but could not integrate. If finally gave up and looked in the back of the book where I found that the answer was given in terms of that integral!
     
  6. Sep 7, 2011 #5
    Re: ODE, bernoulli equation --> leads to crazy integral !

    But how could I get it even in terms of the integral? I need a constant of integration to solve for so y(1) = 1/2
     
  7. Sep 7, 2011 #6

    ehild

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    Re: ODE, bernoulli equation --> leads to crazy integral !

    Substitute z=x/y. You will get an easy separable differential equation.

    ehild
     
    Last edited: Sep 8, 2011
  8. Sep 8, 2011 #7

    ehild

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    Re: ODE, bernoulli equation --> leads to crazy integral !

    With the substitution z=x/y,

    y=x/z, y'=(z-xz')/z^2,the ODE becomes

    x^2(z-xz')/z^2+2x^4/z=(1+2x^2)x^2/z^2.

    Multiplying with z^2/x^2, and rearranging the equation:

    z-xz'+2x^2 z=(1+2x^2), which is a separable ODE:

    xz'=(1+2x^2)(z-1)--->z'/(z-1)=(1+2x^2)/x

    ehild
     
  9. Sep 8, 2011 #8
    Re: ODE, bernoulli equation --> leads to crazy integral !

    ehild, thank you!!

    How do you know to do that substitution?

    thanks again,
    Jonny
     
  10. Sep 8, 2011 #9

    ehild

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    Re: ODE, bernoulli equation --> leads to crazy integral !

    I did not know that substitution, I just tried...

    Dividing both sides of the equation by y^2*x^2, it becomes

    y'/y^2+ 2*(x/y) = 1/x^2+2.

    So why not try to choose x/y the new variable? The derivative of x/y will contain the term y'/y^2...

    There are no recipes for every ODE-s, and there can be different methods for the same ODE. The recipe suggested the substitution z=1/y, but it leads to a difficult integral. You have to be creative :smile:

    ehild
     
  11. Sep 9, 2011 #10
    Re: ODE, bernoulli equation --> leads to crazy integral !

    thanks very much!!!!!!!!

    jonny
     
  12. Sep 9, 2011 #11

    tiny-tim

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    oops!

    oops! :redface:

    your original method was fine :smile:, and on careful inspection your final integral

    ∫ -e-x2(2 + 1/x2) dx​

    is simply e-x2/x + C :blushing:
     
  13. Sep 9, 2011 #12

    ehild

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    Re: oops!

    Ingenious! How can one find out this integral?

    ehild
     
  14. Sep 9, 2011 #13

    tiny-tim

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    just by looking hard enough …

    if it's (fe-x2)',

    then that's (f' -2xf)e-x2, so -2xf = -2 works. ie f = 1/x, f' = -1/x2 :wink:
     
  15. Sep 9, 2011 #14

    ehild

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    Re: ODE, bernoulli equation --> leads to crazy integral !

    You can look really hard. :smile:
    Those who do not have such eyes would input it to Wolframalpha...

    ehild
     
  16. Sep 9, 2011 #15
    Re: ODE, bernoulli equation --> leads to crazy integral !

    It's an initial value problem. So, if you have a linear ODE:
    [tex]
    z' + a(x) z = b(x)
    [/tex]
    and you find the integrating factor:
    [tex]
    \mu(x) = e^{\int{a(x) \, dx}}
    [/tex]
    without caring about the arbitrary integrating constant, because that would give an arbitrary multiplicative factor to your integrating factor, then, the solution to the initial value problem with:
    [tex]
    z(x_{0}) = z_{0}
    [/tex]
    is:
    [tex]
    \mu(x) \, z(x) - \mu(x_{0}) \, z_{0} = \int_{x_{0}}^{x}{\mu(t) \, b(t) \, dt}
    [/tex]

    From here you can immediately see that if [itex]\mu(x)[/itex] gets multiplied by an arbitrary constant, each term in the above expression gets multiplied by it and the equality holds.
     
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