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Introduction to substitutions

  1. Aug 9, 2010 #1
    Hi Guy's,
    I have just started DE's and I'm having difficulty following my notes on using the substitution method. I was wondering if any one new a good link that could show me some good techniques to use when approaching these problems. Is there a good step by step process which any of you use?

    An example in my notes says for

    [itex]y'=\frac{y+2y}{y-sx}[/itex]

    use the substitution [itex]u = \frac{y}{x}[/itex]
    ie

    [itex]y = ux[/itex]

    Is there a reason why they chose [itex]u = \frac{y}{x}[/itex] ?




    regards
     
  2. jcsd
  3. Aug 9, 2010 #2

    Hurkyl

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    Because it works!


    It was probably thought of because y/x appears all over the right hand side, if you reorganize it. (I assume there's a typo in what you wrote?)
     
  4. Aug 9, 2010 #3

    HallsofIvy

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    I'm pretty sure that's incorrect. Wasn't it
    [tex]y'= \frac{y+ 2x}{y- sx}[/tex]

    In any case, if you divide both numerator and denominator of either
    [tex]\frac{3y}{y- sx}[/tex]
    or
    [tex]\frac{y+ 2x}{y- sx}[/tex]
    by x you get
    [tex]\frac{3\frac{y}{x}}{\frac{y}{x}- s}[/tex]
    or
    [tex]\frac{\frac{y}{x}+ 2}{\frac{y}{x}- s}[/tex]
    and taking u= y/x those are
    [tex]\frac{3u}{u- s}[/tex]
    and
    [tex]\frac{u+ 2}{u- s}[/tex]

    Now you have just the one variable, "u" rather than both "x" and "y".

     
  5. Aug 9, 2010 #4
    Sorry it was,

    [itex] y '= \frac{y+2x}{y-2x}[/itex]
     
    Last edited: Aug 9, 2010
  6. Aug 9, 2010 #5
    [itex]y ' = \frac{y+2x}{y-2x}[/itex]

    So like you said if I divide both numerator and denominator by x I get


    [itex]y ' = \frac{\frac{y}{x}+2}{\frac{y}{x}-2}[/itex]

    use the substitution u= y/x


    [itex]y ' = \frac{u+2}{u-2}[/itex]

    now on the left side I still have [itex]y'[/itex]
    ie

    [itex]\frac{dy}{dx} = \frac{u+2}{u-2}[/itex]

    How does the left hand side become a function wrt [itex]\frac{du}{dx}[/itex] ?
     
  7. Aug 9, 2010 #6

    Char. Limit

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    Well, you have y=ux, so differentiate both sides wrt x.
     
  8. Aug 9, 2010 #7
    So do we differentiate [itex]dy/dx dx[/itex]
    and [itex]\frac{u+2}{u-2}dx[/itex]
     
  9. Aug 10, 2010 #8

    HallsofIvy

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    ??? There is no reason to have that additional "dx".

    Since u= y/x, y= xu. Now use the product rule: dy/dx= (dx/dx)u+ x(du/dx)= u+ x du/dx.

    The differential equation becomes
    [tex]x\frac{du}{dx}+ u= \frac{u+ 2}{u- 2}[/tex]

    [tex]x\frac{du}{dx}= \frac{u+ 2}{u- 2}- u \frac{u+ 2- u^2- 2u}{u- 2}= \frac{2- u- u^2}{u- 2}[/tex]

    [tex]\frac{u- 2}{2- u- u^2}du= x dx[/tex]
     
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