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Differential Equations. Reduction of Order. Calc IV.

  1. Sep 13, 2012 #1
    1. The problem statement, all variables and given/known data

    (x2+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0

    2. Relevant equations

    reduction of order, and exact equations..

    3. The attempt at a solution

    let y'=p and y''=p'
    Giving: (x2+2p)p'+2xp Let M=2xp and N=(x2+2p)
    So, Mp=2x and Nx=2x Therefore, they are exact.
    Let fx=2xp
    so, f(x,p)=x2p+h(p)
    fp=x2=h'(p)
    comparing to fp=x2+2p we know h'(p)=2p so, h(p)=p2
    so, c=p2+px2
    Now, solve for p: using quadratic.
    p=(-x2±√(x4+4c))/2
    Substitute back for y'=p
    y'=(1/2)(-x2±√(x4+4c)
    y=?

    This is where i'm not sure what to do.

    I'm pretty sure I use the initial conditions, do I assume c=0 and take the integral of:
    (1/2)(-x2±√(x4))
     
  2. jcsd
  3. Sep 13, 2012 #2

    LCKurtz

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    Right there you have ##y'^2+x^2y'=c##. If you use ##y'(0)=0## you get ##c=0## so your equation is ##y'^2+x^2y'=y'(y'+x^2)=0##. This tells you that either ##y'=0## or ##y'=-x^2##, both of which are easy to handle.
     
  4. Sep 13, 2012 #3
    I'm confused, you're substituting back for p before solving for p? Also, y'(0)≠0 y'(0)=1.

    Does that mean my equation should be:
    y'2+x2y'=1

    And from there solve for y?
     
  5. Sep 13, 2012 #4

    LCKurtz

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    I have quoted your original boundary condition. ##p=y'## so I just put it in. It doesn't matter which you call it, but using ##y'## emphasizes that is what it is, and it makes it more natural when using the boundary conditions.

    No. You get 0 on the right side. It means either ##y'=0## or ##y'=-x^2##. Solve those, apply any boundary conditions and, if you aren't sure, check the solutions work.
     
  6. Sep 13, 2012 #5
    Wow. I got it mixed up with the problem out of the book. It's been a long day. You're right..

    So all I have to do is:

    y=∫0dx=c

    y=∫(-x2)dx=(-x3)/3+c

    Therefore, y=∫(-x2)dx=(-x3)/3+c is the answer?

    Thanks for helping me by the way.
     
  7. Sep 13, 2012 #6

    LCKurtz

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    Use the boundary condition to evaluate c. A second order equation with two initial conditions won't have any arbitrary constants left. And you shouldn't use the same c for both constants.
     
  8. Sep 13, 2012 #7

    LCKurtz

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    I should add here that this is a non-linear equation. You get two separate solutions. Nothing says that the sum of those two solutions is also a solution.
     
  9. Sep 13, 2012 #8
    Oh, so now I say because y(0)=1 0/3+c2=1 so, c2=1?

    Therefore, y=-x3/3+1

    What do you mean don't use the same constant? Are you referring to y=∫0dx=c1?

    What do I do with the y=∫0dx=c1 anyway?
     
  10. Sep 13, 2012 #9

    LCKurtz

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    You apply the boundary condition ##y(0)=1##. Remember you used ##y'(0)=0## to get the two separate solutions. Each of them satisfies ##y(0)=1##. So you have two different solutions to this problem. You should check that they both work.
     
  11. Sep 14, 2012 #10
    Ya, I figured that out after asking. But how do I check that both -x3/3+1 and 1 work?
     
  12. Sep 14, 2012 #11

    LCKurtz

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    Plug them into the equation and boundary conditions.
     
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