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Homework Help: Differential Equation substitute-to-make-seperable question

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    Problem #7 in this old exam: http://math.njit.edu/pdfs/222ex1.pdf [Broken]

    ... note: i didnt care about getting the IVP solvied... this is just practice and getting the general solution was all i cared about doing

    2. The attempt at a solution
    http://wolfsfiles.googlepages.com/scan.jpg

    I think i confirmed my answer in a cas. But my confussion is if i plug the original differential equation, without making that substitution, that i get a differant general solution... and i'm not entirely sure if thats to be expected.

    In short, is my answer correct :P

    3. Quick question about a differant problem
    Problem #3-A on that old exam:
    "(d/dx)[sec(x)y]=x ...get general solution"

    should the left side be expanded first, and then work the problem, or am i not seeing some easy way to seperate this thing right away.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 24, 2007 #2

    rock.freak667

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    Homework Helper

    Well when i did the first one I eventually got

    [tex]\frac{1}{2x}v' = \frac{v^2+1}{v}[/tex]

    and I believe it was because of this step

    [tex]y=vx[/tex] so [tex]\frac{dy}{dx}=x\frac{dv}{dx} + v [/tex]


    you missed out the extra v and put 0 instead

    and for [tex]\frac{d}{dx}{[sec(x)y]=x [/tex]

    I think they are telling you that the differential of sec(x)y w.r.t x is equal to x....so you can just integrate both sides w.r.t x
     
  4. Sep 25, 2007 #3
    Thanks for the help. Teacher confirmed how you did the substitution, but think we got differant answer, or its in a differant form.

    From just sub-ing in y = vx the general solution from that was
    v^2 = CX-1, which was de-substituted to become
    (Y/X)^2 = CX - 1 by replacing v with y/x

    and y(1) = 1 was the initial condition... ya i know i ddidnt want to do the ivp,, but this is answer anyhoot

    gives... C = 2

    so exact solution is: (y/x)^2 = 2x -1

    so ya my problem was i differentiated wrong... was trying to do partial derivitives but i think the goal was implicit differentation... and it got the respect-to-x from the original differential equation which i didn't realize to do. Thx for the help.
     
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