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Differential Equation- Separation of Variables

  1. May 19, 2008 #1
    1. The problem statement, all variables and given/known data
    y'=xsec^2(x^2)

    2. The attempt at a solution
    dy/dx=xsec^2(x^2)
    dy=xsec^2(x^2)dx
    [tex]\int[/tex]dy=[tex]\int[/tex]xsec^2(x^2)dx
    lny= (here i'll do a u substitution)
    ----
    u=x^2 du=1/3x^3dx


    ... and here's my problem. It seems like that creates a very difficult u-sub to try and manage. Any suggestions/help?
     
  2. jcsd
  3. May 19, 2008 #2

    Defennder

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    This is incorrect. Your method is correct, though.
     
  4. May 19, 2008 #3
    I think you should write sec function in terms of cos. Function will be dy = x / (cos^2(x^2)). Then if you say x^2 = u and du = 2x.dx, you can solve the problem easily.
     
  5. May 19, 2008 #4
    u=x^2 du=2xdx
    2[tex]\int[/tex]du/cos^2(u)
     
  6. May 19, 2008 #5

    Defennder

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    You're off by a factor of 1/4. It should be [tex]\frac{1}{2} \int sec^2u \ du[/tex].
     
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