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Integration Help

  1. Dec 10, 2006 #1
    Simple Diff Eq Help

    I am trying to solve the following Diff Eq:

    [tex]\frac{d^2x}{dy^2}+(\frac{y}{2}-\frac{1}{y})\frac{dx}{dy}=0[/tex]

    I tried to solve by setting [tex]\frac{dx}{dy}=z[/tex]

    so: [tex]\frac{dz}{dy}+(\frac{y}{2}-\frac{1}{y})z=0[/tex]

    I know the general solution to this is:

    [tex]z=-e^{-\int{\frac{y}{2}-\frac{1}{y}dy}}\int{0}dy[/tex]

    This then yields:
    [tex]z=-C_1e^{ln(y)-1/4y^2}=\frac{dx}{dy}[/tex]

    And trying to integrate again, Using u-substitution, [tex]u = ln(y) -1/4y^2[/tex]
    [tex]du=(\frac{1}{y} - \frac{y}{2}) dy[/tex]
    [tex]dy = \frac{2}{2-y^2} du[/tex]

    Now, can I leave that y tern in the u-substitution? Or did I make a mistake along the way?

    Also, is there an easier way to solve this integral than the path I've taken? Thanks.
     
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2
    I'd try not thinking about what the general solution should be and think about seperating the equation and solving from there.

    however for you final integral, you need to symplify z a bit.

    the integrand should be

    Cye^-1/4y^2

    which is easily integrable using the substitution u=y^2
     
  4. Dec 11, 2006 #3

    HallsofIvy

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    Staff Emeritus
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    That's valid, since this is a linear equation but I think it would be simpler to treat it as a separable equation:
    [tex]\frac{dz}{z}= \left(\frac{1}{y}- \frac{y}{2}\right)dy[/itex]
    so
    [tex]ln(z)= ln(y)- \frac{y^2}{4}+ c[/tex]
    or
    [tex]z= \frac{dx}{dy}= \frac{Cy}{e^{\frac{y^2}{4}}}[/tex]
    That also separates:
    [tex]dx= \frac{Cy}{e^{\frac{y^2}{4}}}dy[/tex]
    To integrate that, let u= y2/4.

    No, you cannot leave "y" in something you are integrating with respect to u.

     
    Last edited: Dec 13, 2006
  5. Dec 12, 2006 #4
    Thanks for the help!
     
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