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Help nedded to solve a D.E. using substitution

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the appropriate substitution to solve the following D.E.: -ydx + (x + [tex]\sqrt{}xy[/tex])dy = 0

    2. Relevant equations

    y = ux

    3. The attempt at a solution

    y = ux implies dy = udx + xdu

    so -xudx + (x + x[tex]\sqrt{}u[/tex])(udx + xdu) =0

    we then get after some simplificaion: xu[tex]\sqrt{}u[/tex] dx + x2 (1 + [tex]\sqrt{}u[/tex])du = 0

    so (1/x).u[tex]\sqrt{}u[/tex]dx + (1 + [tex]\sqrt{}u[/tex])du = 0

    hence dx/x + du/(u[tex]\sqrt{}u[/tex]) + du/u = 0

    now we have after integrating: lnx + lnu - 2/[tex]\sqrt{}u[/tex] = c1

    substituting bk u= y/x we have: ln x + ln (y/x) - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c1

    ln x + ln y - lnx - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c1

    so ln y - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c1

    here i got stuck. i couldn;t continue although i know that the answer should be : 4x = y(ln|y| - c)2

    need help in this plz. my process is right isn't it? how should i continue?
     
  2. jcsd
  3. Nov 16, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The rest is just basic algebra!
    From [itex]ln|y| 2\sqrt{x}/\sqrt{y}= c1[/itex], [itex]2\sqrt{x}/sqrt{y}= ln|-|c1[/itex],
    [itex]2\sqrt{x}= (ln|y|- c1)\sqrt{y}[/itex]. Squaring [itex]4x= (ln y- c1)^2y[/itex].
     
  4. Nov 16, 2008 #3
    thank u Sir for the help.

    I feel like a dum after it appeared to be that apparent.

    thx again
     
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