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Homework Help: Homogenous ode problem

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data
    this i a problem on homogeneous ode given as an exercise in class
    [tex] \frac {dy} {dx} = \frac {2xy} {x^2+y^2} [/tex]
    im supposed to find the solution,
    2. Relevant equations
    substitute [tex] v= \frac {y} {x} [/tex]
    do it as a homogeneous ode and on and on

    3. The attempt at a solution
    i worked all the way to this point
    [tex] \int \frac {dv} {dx} = \int \frac {v+v^3} {1+v^2} [/tex] after the trivial substitution for homogeneous odes and stuff [itex] v= \frac {y} {x} [/itex]
    which then i tried separation of variables on it,
    [tex] \int \frac {1+v^2} {v+v^3}dv = \int \frac {dx} {x} [/tex]
    the right side is trivial so i will not continue to blabber about it,
    now separating the left into two equations ;
    [tex] \int \frac {1} {v+v^3} dv +\int \frac{v^2} {v+v^3} dv =... [/tex]
    the first integral there has me stumped the second well seems doable by substitution, unless its more deceptive than i think since i got an ans of [itex] \frac {1} {2} \ln (1+v^2) [/itex] after doing the substitution [itex] u=1+v^2 [/itex] , who can help me with that first integral [itex] \int \frac {1} {v+v^3} dv [/itex] ?
    Last edited: Sep 5, 2009
  2. jcsd
  3. Sep 5, 2009 #2


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    Welcome to Physics Forums.

    For your first integral, try taking a factor of v out of the denominator and then using the substitution v = tan(theta).
  4. Sep 5, 2009 #3
    thanks man,
  5. Sep 5, 2009 #4


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    Yes, but why? In addition to being "homogeneous", that de is separable and it is easier doing it that way. In fact, it is probably a good idea to recognize immediately that, for [itex]x\ne 0[/itex],
    [tex]\frac{2xy}{x+ x^3}= \frac{2y}{1+ x^2}[/tex]

    So this "separates" as
    [tex]\frac{dy}{y}= 2\frac{dx}{x^2+ 1}[/tex]
    and both sides are easy to integrate.
  6. Sep 5, 2009 #5
    you ask why?
    because its a class exercise for the method :) , and i realize i wrote the wrong question but the working for the right question so here is the correct question [tex] \frac{dx}{dy}=\frac{2xy} {x^2+y^2}[/tex] i will correct the posting too
  7. Sep 5, 2009 #6

    is this what you mean? : [tex] \frac{1}{v(1+v^2)} [/tex] and substituting [itex] v=tan(\theta) [/itex]
    [tex] dv= tan^2(\theta) + 1 [/tex] now we have [tex]\int \frac{tan^2(\theta) + 1 }{tan\theta (1 + tan^2 (\theta))} d\theta [/tex] after cancelling :
    [tex] \int \frac {d\theta} {tan(\theta)} [/tex] but tan is identical to [itex] \frac{sin}{cos} [/itex] using [itex] u=sin(\theta) [/itex] we got now: [tex] \int \frac{du} {u} [/tex] and the grand finale [tex] \ln(sin(\theta)) [/tex]
  8. Sep 5, 2009 #7


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    That is indeed what I mean. However, you should note that this integral no longer applies to your "corrected" question.
  9. Sep 5, 2009 #8
    why? The working i gave was for the corrected question , not the first erroneous one
  10. Sep 5, 2009 #9


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    Starting from your corrected ODE,

    [tex]\frac{dy}{dx} = \frac{2xy}{x^2 + y^2}[/tex]

    and substituting y=vx yields,

    [tex]x\frac{dv}{dx} = \frac{2v}{1+v^2} - v[/tex]

    [tex]x\frac{dv}{dx} = \frac{v - v^3}{1+v^2}[/tex]

    [tex]\int\frac{1+v^2}{v\left(1-v^2\right)}\;dv = \int\frac{dx}{x}[/tex]

    Which does not match your integrals. It seems that you have dropped a sign.
  11. Sep 6, 2009 #10
    thanks, my signs need some more work,
    from here is this how this would be done: [tex]\int\frac{1}{v\left(1-v^2\right)}\;dv [/tex]
    let [itex] v= sin\theta [/itex] , [itex] dv= cos\theta d\theta[/itex] and we have,
    [tex] \int \frac {cos\theta}{sin\theta cos^2 \theta }d\theta[/tex]
    [tex] 2\int \frac {1}{sin2\theta}d\theta [/tex] and
    [tex] 2\int csc2\theta d\theta [/tex]
  12. Sep 6, 2009 #11


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    From here
    note that

    [tex]\sin2\theta = \frac{2\tan\theta}{1+tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta}[/tex]


    [tex]2\int \frac {1}{sin2\theta}d\theta = \int\frac{\sec^2\theta}{\tan\theta}d\theta[/tex]

    And note that,

    [tex]\frac{d}{d\theta}\tan\theta = \sec^2\theta[/tex]

    From here, your integral is trivial.
  13. Sep 6, 2009 #12


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    \frac{1+v^2}{v\left(1-v^2\right)}\ = \frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}

  14. Sep 7, 2009 #13
    Thanks, i get the method now
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