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Second Order Differential Equation (homogenous)

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Given [itex]y_1(x)=x[/itex] is a solution to [itex](2x-1)y''-4xy'+4y=0[/itex], find [itex]y(2)[/itex] given [itex](y(1),y'(1))=(0, 0)[/itex]. Utilize method of reduction of order.

    I need help with this as I end up getting some ugly (in my mind, anyways) integrals. Thanks in advance!


    3. The attempt at a solution

    Let [itex]y=y_1v=xv[/itex], hence [itex]y'=v'x+v[/itex], and [itex]y''=2v'+v''x[/itex].

    [tex]4xv'-2v'+2x^2v''-xv''-4xv-4x^2v'+4xv=0[/tex]
    [tex]4xv'-2v'+2x^2v''-xv''-4x^2v'=0[/tex]
    Let [itex]w=v'[/itex]
    After separating I get:
    [tex]\frac{dw}{w}=\frac{4x^2-4x+2}{2x^2-x}dx[/tex]

    [tex]\frac{4x^2-4x+2}{2x^2-x}=2+\frac{A}{x}+\frac{B}{2x-1}[/tex]
    where [itex] A(2x-1)+Bx=-2x+2[/itex], hence [itex](A, B)=(-2, 2)[/itex]
    Therefore,
    [tex]\frac{dw}{w}=(2-2\frac{1}{x}+2\frac{1}{2x-1})dx[/tex]
    [tex]ln|w|=2x-2ln|x|+2ln|2x-1|+C_1[/tex]
    [tex]w=\frac{e^{2x+C_1}(4x^2-4x+1)}{x^2}[/tex]
    Next, I know I am supposed to integrate this to determine y', however, I am unsure of how to go about doing this.
     
    Last edited: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2

    haruspex

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    You should be able to determine C1 at this stage from the initial conditions. The answer is rather interesting, but I'm not sure what the implications are.
     
  4. Feb 26, 2013 #3
    Well, [itex]w=v'=\frac{y'-v}{x}=\frac{y'-y/y_1}{x}[/itex], however, the initial values causes this to fall to zero, so I am a little confused.
     
  5. Feb 26, 2013 #4

    haruspex

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    Exactly. A degenerate solution, perhaps?
     
  6. Feb 26, 2013 #5
    My apologies. This is the first time I have seen this, and my textbook does not seem to mention anything regarding degenerate solutions. What exactly does this imply? Does that mean I assume [itex]v'=0[/itex] and start again?
     
  7. Feb 26, 2013 #6

    haruspex

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    It implies to me that with these initial conditions y will be always 0. That's certainly a solution.
     
  8. Feb 26, 2013 #7
    Okay, let's say that y(1)=1 instead.
    That results in [itex]-1=e^{2x+C_1}[/itex] which is impossible, however, the example that I am looking at states that if y(1)=1 there is a solution. What sort of trick does one do after getting to this point, though? If it were zero I can see how that would indicate that v was a constant since v'=0. However, being that it is equal to -1 with y(1)=1, y'(1)=0, I don't quite see what that implies.
     
    Last edited: Feb 26, 2013
  9. Feb 26, 2013 #8

    haruspex

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    No, you correctly had ln|w| on the LHS, not ln(w).
     
  10. Feb 26, 2013 #9
    Ah yes, don't know how that got away from me. Which brings me to [itex]C_1=-2[/itex]. I still have to do a nasty integral to determine [itex]v[/itex] do I not?
     
  11. Feb 26, 2013 #10

    haruspex

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    I believe that integral can be reduced/equated to integrating a double exponential. There's nothing special about the range, so it cannot be done analytically.
    But how come you changed the b.c. from y(1)=0 to y(1)=1? Which is actually wanted?
     
  12. Feb 26, 2013 #11
    There was both that I wanted to go over. Thanks for your help! Now that I know I cannot analytically solve that, I am done.
     
  13. Feb 27, 2013 #12
    I have been notified that there is an error as this problem is indeed doable analytically. Here is my adjusted work, however, still seems rather tricky to solve:
    Let [itex]y=y_1v=xv[/itex], hence [itex]y'=v'x+v[/itex], and [itex]y''=2v'+v''x[/itex].

    [tex]4xv'-2v'+2x^2v''-xv''-4xv-4x^2v'+4xv=0[/tex]
    [tex]4xv'-2v'+2x^2v''-xv''-4x^2v'=0[/tex]
    Let [itex]w=v'[/itex]
    After separating I get:
    [tex]\frac{dw}{w}=\frac{4x^2-4x+2}{2x^2-x}dx[/tex]

    [tex]\frac{4x^2-4x+2}{2x^2-x}=2+\frac{A}{x}+\frac{B}{2x-1}[/tex]
    where [itex] A(2x-1)+Bx=-2x+2[/itex], hence [itex](A, B)=(-2, 2)[/itex]
    Therefore,
    [tex]\frac{dw}{w}=(2-2\frac{1}{x}+2\frac{1}{2x-1})dx[/tex]
    [tex]ln|w|+c_1=2x-2ln|x|+ln|2x-1|[/tex]
    From here we can utilize [itex]y'=v'x+v[/itex], since [itex]w=v'[/itex] we have [itex]y'(1)=w(1)(1)+1[/itex] and [itex]w(1)=-1[/itex].

    [tex]ln|-1|+c_1=2,c_1=2[/tex]
    Therefore,
    [tex]v=\int\frac{e^{2x-2}(2x-1)}{x^2}dx[/tex]

    From here you can get the right answer by factoring out e^(-2) and applying integration by parts.
     
    Last edited: Feb 27, 2013
  14. Feb 27, 2013 #13

    haruspex

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    I got that far before, but it didn't seem like integration by parts would help. What I didn't consider was whether the nasty parts might cancel, so here goes:
    ##\int_1^2\frac{e^{2x}(2x-1)}{x^2}dx = 2\int_1^2\frac{e^{2x}}{x}dx - \int_1^2\frac{e^{2x}}{x^2}dx = 2\int_1^2\frac{e^{2x}}{x}dx + [\frac{e^{2x}}{x}]_1^2 - 2 \int_1^2\frac{e^{2x}}{x}dx##
    Bingo.
     
  15. Feb 27, 2013 #14
    [tex]e^{-2}[2\int\frac{e^{2x}}{x}dx-\int\frac{e^{2x}}{x^2}dx]=e^{-2}[2\int\frac{e^{2x}}{x}dx-\frac{e^{2x}}{x}-2\int\frac{e^2x}{x}dx]=-\frac{e^{2x-2}}{x}+c_2[/tex]
    I used [itex]U=e^{2x}, dU=2e^{2x}dx, dV=1/x^2dx, V=-1/x[/itex]

    From here I determined [itex]c_2=2,[/itex], and [itex]y(2)=-3.38906[/itex] which was the solution.
     
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