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Method of reduction of order

  1. Jun 12, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Solve the following using the method of reduction of order.

    [itex]4x^2y''+y=0[/itex]
    [itex]y_1=\sqrt{x}lnx[/itex]


    2. Relevant equations

    [itex]y_2=v(x)y_1[/itex]

    v(x) = ∫ (e^(-∫P(x)dx / y_1) Sorry if that's tough to read, couldn't figure out the latex for it, but it's supposed to be a forumla for V(x).

    3. The attempt at a solution

    K so, Y_2 = V(x)y_1.

    In this case P(x) is 0, correct? Because there is no y' term.

    That's what is throwing me off. Our teacher often just sort of ignores constants when he's teaching, so I'm not quite sure what to do.

    I end up with

    [itex]V(x) = \int e^k x^{-1/2} lnx dx[/itex]

    I don't know what to do with that. I'm not sure if he just wants us to say that V(x)y_1 is the solution, or if he wants us to simplify it out.

    Any thoughts? Either way I wouldn't have the slightest clue how to go about integrating that.
     
  2. jcsd
  3. Jun 12, 2012 #2

    LCKurtz

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    I don't know what the ##k## in your answer is and without you showing more steps it's hard to help you. Your ##V(x)## isn't correct. Don't forget you need the leading coefficient to be ##1## when you calculate the integrating factor. Show us your work.

    For what it's worth, the second solution comes out nice and simple.
     
  4. Jun 12, 2012 #3

    HallsofIvy

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    Since you didn't tell us what "P(x)" is supposed to mean, we cannot tell you what function it is. As for "[itex]V(x)= \int e^k x^{1/2}ln(x)dx[/itex]", where did that "k" come from? There is no "k" in anything you wrote before.
     
  5. Jun 13, 2012 #4

    ElijahRockers

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    P(x) is the coefficient of the y' term. So P(x) is zero.... the integral of 0 is K, that's where the K came from.

    When you say V(x) is incorrect, do you mean the formula or the V(x) I ended up with? The latter wouldn't surprise me, that's why I'm here for help, but the former I copied from the board.

    If I go through the steps to solve for V(x) the problem gets very messy because there are so many derivatives of x^(1/2)ln x. That's why I assumed he wanted us to use the formula.

    As for more steps, well... according to my teacher I should just plug P(x) and y1 into the formula for V(x) and multiply by y1 to get y2.

    Anyway, I am about to head to class and I will ask him. I will post my findings here.

    Thanks for the help.
     
    Last edited: Jun 13, 2012
  6. Jun 13, 2012 #5

    LCKurtz

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    Just to give you a point of reference, if you go through the steps to find ##v(x)##, you should wind up with this DE after simplifying$$
    v''x\ln x+v'\ln x+2v'=0$$And don't forget you need the leading coefficient to be 1 when you calculate the integrating factor so you can solve for ##v'##.
     
  7. Jun 13, 2012 #6

    ElijahRockers

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    I got some help today on campus. The problem was with my formula for V(x). I have it written in my notes wrong. The actual formula should have a denominator of y^2, not just y.

    Also about the constant, the teacher told me just to assume it is 0. I will probably never take another 5-week math course, it is making my head spin.

    Everything fell into place once I realized that, but thank you both for your help!
     
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