1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook