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Second order substitution!

  1. Aug 24, 2006 #1
    I am having real trouble with this second order differential
    The substitution is given and i just cant seem to use it
    What am i missing here?

    [tex] x \frac{d^2 y} {dx^2} -2 \frac{dy} {dx} + x = 0, \frac{dy} {dx} = v [/tex]

    All help welcome
     
  2. jcsd
  3. Aug 24, 2006 #2
    You need to show some work. What problems are you having with that substitution, it shouldn't be very hard to use.
     
  4. Aug 25, 2006 #3
    That's just the problem. I can't get much at all.
    The only step I can think of is:

    [tex] x \frac {dv}{dx} + 2v + x = 0 [/tex]

    I don't see where that leads me.
    Do i need to find x in terms of v?...
     
  5. Aug 25, 2006 #4
    From where you are put it in standard form and then think about how you would solve that ODE.
     
  6. Aug 25, 2006 #5
    I've honestly tried for ages to put it in standard form.
    I've tried finding dy/dv, dx/dv, dv/dy etc...
    None of that has worked at all...
     
  7. Aug 25, 2006 #6
    No you need to find v in terms of x. Solve that relatively simple ODE for the function v and then you have another even simpler equation that if you solve will give you y.

    you have

    xv' + 2v + x = 0

    so

    xv' + 2v = -x

    Can you solve that equation?
     
  8. Aug 26, 2006 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In your first post, you said the equation was
    [tex] x \frac{d^2 y} {dx^2} -2 \frac{dy} {dx} + x = 0, \frac{dy} {dx} = v [/tex]
    Now, after substituting v for dy/dx, you write
    [tex] x \frac {dv}{dx} + 2v + x = 0 [/tex]

    One of those has a typo. Assuming the first equation is what you want, the second should be
    [tex] x \frac {dv}{dx} - 2v + x = 0 [/tex]
    or, as d leet said, xv'- 2v= -x (correcting that sign error).
    That's a linear first order differential equation and has a simple integrating factor.

    Another method: write the equation as xv'= 2v- x and divide by x:
    v'= 2(v/x)- 1. Now let u= v/x so that v= xu. v'= xu'+ u and the equation becomes xu'+ u= 2u- 1 or xu'= u-1, a simple separable equation.
     
  9. Aug 26, 2006 #8
    Ahhh thank you...
    I couldn't see what i was missing
    It's all so simple :blushing:
     
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