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Differential Equations

  1. Jan 15, 2007 #1
    Hey all,

    I have an exam coming up and the pattern of questions seems to include some that I am stumped with.

    One goes like this:

    The second order differential equation

    x2y'' - 6xy' + 12y = 0,​

    has non-constant coefficients. Find the general solution by looking for a solution y = xn and finding a quadratic equation for n.

    I would love to say "this is what I have done" but I just couldn't see anything. I tried substituting y = xn, y' = nxn-1 and y'' = n(n-1)xn-2 into the equation but that just doesn't help, so far as I can see. I need a nudge in the right direction.

    Cheers,

    The Bob (2004 ©)
     
    Last edited: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2

    cristo

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    What do you get when you substitute y=xn (and derivatives) into the equation? Why do you say that this won't help? I think that's the correct way to approach this question!
     
    Last edited: Jan 15, 2007
  4. Jan 15, 2007 #3

    Integral

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    Show us what you get after making the substitutuons for

    [tex] y = x^n [/tex] and its derivatives.
     
  5. Jan 15, 2007 #4
    Ok, well... after substitution and messing around with powers, which is probably why I didn't see it going anywhere (at least), I get:

    xn(n - 4)(n - 3) = 0​

    I am assuming I would be correct to say that n = 4 and n = 3.

    So y = x3 + x4 with the quadratic for n as (n - 4)(n - 3) = 0

    How much of this is good and how much can you tell is guess work?

    Cheers,

    The Bob (2004 ©)
     
    Last edited: Jan 15, 2007
  6. Jan 15, 2007 #5
    Wouldn't you just leave it as "y = x^4, y = x^3"?
     
  7. Jan 16, 2007 #6

    Mute

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    It's all correct. When you sub in y = x^n, since the differential equation has the form [itex]ax^2y'' + by' + cy = 0[/itex], the reduction in the power due to differentiations on [itex]x^n[/itex] are cancelled out by the x^2 in front of y" and the x in front of y'. You can thus factor out the [itex]x^n[/itex] to get [tex]x^n(an(n-1) + bn + c) = 0[/itex], which you know is only true for all possible values of x if the quadratic in n is equal to zero. In the specific problem you mentioned, that's why n = 3, 4.

    And so, you get the solutions y = x^4 and y = x^3, and your general solution is a linear combination of them, [itex]y = Ax^3 + Bx^4[/itex].

    So it appears you knew what you were doing, you just weren't sure of it.
     
  8. Jan 16, 2007 #7

    HallsofIvy

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    You don't need to "assume" it. This equation must be true for all x. In particular for x= 1 which means (n-4)(n-3)= 0.

    Youwere good up to this point! Any solution to a second order, linear, homogeneous differential equation can be written as a linear combination of two independent solutions. x3 and x4 are independent solutions. What does a linear combination of them look like?
     
  9. Jan 16, 2007 #8
    Yeap, this entire step eluded me. Well it didn't but and it is annoying because when I looked at it first time I thought about doing somthing like this.

    y = Ax3 + Bx4, although I have worked it through and it would appear that A and B are can be any constant.

    Anyway cheers all,

    The Bob (2004 ©)
     
  10. Jan 16, 2007 #9
    Whilst I am on a role, I have another question before tomorrow.

    Given that y = ex is one solution of the differential equation

    (1 + x)y'' - (1 - 2x)y' + (x - 2)y = 0​

    find the general solution.

    What I have done is let y = uex, differentiate this and substitued in. Then I have let w = u' and got down to:

    w'(ex + xex) + 3exw = 0​

    The problem from here is that the way I have been told to solve it is to assume it is in the form:

    y' + P(x)y = Q(x)​

    which does not allow a solution, unless, again, I am doing it wrong.

    Any pointers?

    Cheers,

    The Bob (2004 ©)
     
  11. Jan 17, 2007 #10

    Mute

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    Keep in mind that e^x is never zero, so you can divide it out of your equation in w to get

    [tex](1+x)w' + 3w = 0[/tex],

    which is a linear first order differential equation that you can solve using an integrating factor.
     
  12. Jan 17, 2007 #11

    dextercioby

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    I'd say it's better if you simply separate variables and integrate instead of calculating the integrating factor.

    Daniel.
     
  13. Jan 17, 2007 #12
    May I thank you both. Although I do not understand completely why you can simply divide through by ex because it is never zero, I can appreciate what is being said and, unfortunately for now, learn that bit parrot fashion.

    After this division, I could see where the seperable equation was with more ease.

    Cheers again :biggrin:

    The Bob (2004 )
     
  14. Jan 17, 2007 #13

    HallsofIvy

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    One serious problem you have with this is that ex is NOT a solution to that equation!

    Apparently, from what you have below the equation is actually
    (1+ x)y"+ (1- 2x)y'+ (x-2)y= 0.
     
  15. Jan 17, 2007 #14

    Mute

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    What if instead of e^x you had had [itex](2+2x)w' + 6w = 0[/itex] or [itex](117 + 117x)w' + 351w = 0[/itex]? In each case you would multiply both sides of the equation by 1/2 and 1/117, respectively, to get rid of the common factor in each term. The same goes for e^x: You're multiplying both sides of the equation by 1/e^x = e^(-x) to remove those terms from your equation. Equivalently, you could just factor it out, like you do with the [itex]x^n[/itex] in your first problem, to get [itex]e^x((1+x)w' + 3w) = 0[/itex] and note that for this equation to hold for all x, you need [itex](1+x)w' + 3w = 0[/itex].
     
    Last edited: Jan 17, 2007
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