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A relatively simple ODE problem

  1. Jan 28, 2010 #1
    So many years ago, i took elementary differential equations. unfortunately, i've forgotten too much of it, and now i'm in a not so elementary class. i'm confident i'll catch up to where i was, but at the moment i'm stuck on a relatively simple problem.


    i'm having trouble separating them (not sure if this is separable). maybe if someone could give me a pointer to get started, i would be very appreciative. thanks!
    1. The problem statement, all variables and given/known data


    x3 + 3y -xy' = 0.

    2. Relevant equations



    3. The attempt at a solution
    stuck
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 28, 2010 #2

    Dick

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    It doesn't look separable to me. Try finding an integrating factor. Does that ring a bell?
     
  4. Jan 29, 2010 #3
    okay, so i'm not getting the correct answer, but here's the work i've done so far:
    x3 + 3y -xy' = 0
    3y - xy' = -x3
    x(3y/x) - y' = -x3
    -y' + 3/x*y = -x2

    let mu(x) = eintegral(-3/x) = e-3 ln x = 1/x3

    d/dx[1/x3*y] = 1/x3* x2
    1/x3*y = x-1
    y=x2

    is this the right path?

    editing:
    i saw i didn't integrate the left hand side at x-1. afterwards, it seemed to work.
    y = x3ln(x)
     
    Last edited: Jan 29, 2010
  5. Jan 29, 2010 #4

    vela

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    Almost. You forgot the constant of integration, which will result in another term, cx3, which is the solution to the homogeneous equation.
     
  6. Jan 29, 2010 #5
    cool, thanks so much.
     
  7. Jan 29, 2010 #6
    it's slowly coming back to me. okay, linear equations, put them in the proper form, use integrating factor, got it.
    now i'm kind of stuck again with a problem from the same set.

    xy + y2 - x2y'=0.

    so this isn't separable, and it's not linear, and i'm kind of stuck. any help, or advice where to start looking? gracias.
     
  8. Jan 29, 2010 #7

    Dick

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    If you divide both sides by x^2, you've got y/x+(y/x)^2-y'=0. Suggests that a change of variable v=y/x might help.
     
  9. Jan 29, 2010 #8
    how's this:

    [tex]\frac{y}{x}[/tex] + ([tex]\frac{y}{x}[/tex])2 - y' = 0
    let v= [tex]\frac{y}{x}[/tex], with [tex]\frac{dy}{dx}[/tex] = v + x[tex]\frac{dv}{dx}[/tex]

    v + v2 - v - x[tex]\frac{dv}{dx}[/tex] = 0
    v2 - x[tex]\frac{dv}{dx}[/tex] = 0

    [tex]\frac{1}{v2}[/tex] dv = [tex]\frac{1}{x}[/tex]dx

    integrate, get

    v = [tex]\frac{-1}{ln x}[/tex]

    substitute and solve for y

    y = x ([tex]\frac{-1}{ln x}[/tex]

    and do i just throw a constant on at the end?
     
  10. Jan 29, 2010 #9

    Dick

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    It's pretty good. But you don't just "throw a constant on at the end". You put in a constant at the point you integrate then carry it through.
     
  11. Jan 30, 2010 #10
    yeah, i kind of figured that, it just seems that that's how it usually works out, and i can't remember to put them in when they come up.
    so i'm having another hard time working out a different problem.

    (for future reference, should i start a new thread if i have a different problem?)

    so it's a bernoulli equation
    4xy2 + y' = 5x4y2
    y' = 5x4y2 - 4xy2
    y' = y2( 5x4 - 4x)

    let n = 2, p(x) = 0, g(x) = 5x4 - 4x, v = y-1
    divide both sides by y2
    y-2y' + 0y = 5x4 - 4x
    substitute v= y-1

    -dv/dx + 0v = 5x4 - 4x
    let mu(x) = eint. 0 dx = ec0 = c1
    -c1 dv/dx + 1*0v = c1(5x4 - 4x)
    d/dx c1*v = c1(5x4 - 4x)
    integrate both sides
    c1*v=-c1*x2(x3 + 2)
    v = -x2(x3 + 2)
    replace v
    y-1 = -x2(x3 + 2)
    y = 1/ (-x2(x3 + 2))
    however, this doesn't seem to work. i'm guessing maybe letting p(x) = 0 might be a problem? i don't see why, it's continuous, but that's the only thing i can think of.

    btw, thanks for your help. it's coming back to me. i was able to solve 2 other bernoulli equations and got stuck on this one.
     
  12. Jan 30, 2010 #11

    vela

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    You're making things hard for yourself. You should be able to see at this point that the equation is separable.
     
  13. Jan 30, 2010 #12

    vela

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    You dropped a sign or two, and again, you forgot the constant of integration.
     
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