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Homogenous ode

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data
    the homogenous ode
    [tex] (x^2+y^2)dy-y^2 dx =0 [/tex]

    2. Relevant equations

    [itex] v=\frac{y}{x} [/itex]

    3. The attempt at a solution
    worked al the way here [tex] \int \frac {1}{v^2-v(1+v^2) }dv + \int\frac{v^2}{v^2- v(1+v^2)} = \int \frac {dx}{x} [/tex]
    how do i handle the integrals ?
  2. jcsd
  3. Sep 11, 2009 #2


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    I would first put it back together again:
    [tex]\int\frac}(1+ v^2)dv}{v^2- v- v^3}= -\int\frac{(1+ v^2)dv}{v(v^2- v+ 1)}[/tex]
    and, since it is a rational integral, use partial fractions.
  4. Sep 12, 2009 #3
    The integral i get after getting partial fractions has me beat: i got A=1, B=-1
    [tex] \int\frac {1}{-v}dv - \int\frac {1}{v^2-v+1} =\int \frac{dx}{x} [/tex]
    the second integral on the left is quadratic and irriducible how is it integrated?
  5. Sep 12, 2009 #4
    Im thinking about completing the square thus the second integral on the left becomes:
    [tex] - \int \frac{dv} {(v-\frac{1}{2} )^2 + \frac{3}{4} }[/tex]
    Last edited: Sep 12, 2009
  6. Sep 12, 2009 #5


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    Yes, that's correct. An irreducible quadratic term can always be written as [itex](v- a)^2+ b[/itex] and then the substitution [itex]u= (v- a)/\sqrt{b}[/itex] reduces it to [itex](1/b)(1/(u^2+ 1))[/itex] and the integral is an arctangent.
    Last edited by a moderator: Sep 12, 2009
  7. Sep 12, 2009 #6
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