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Math Help - desperate seconds

  1. #1
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    Red face desperate seconds

    I have 3 questions and NO TIME.

    I'll be sucking on one of the exam desks just a few hrs later.

    Pls help me coz late studies may bring me just a few points...

    1) xy'' + xy' - y = xlnx maybe about euler


    2) y'' – 2y' + y = x- ex


    3) ty''+(1-2t)y'-2y=0 y(0)=1 , y'(0)=2 Laplace


    Pls help me
    Last edited by mr fantastic; December 17th 2010 at 09:28 PM. Reason: Restored deleted questions.
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  2. #2
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    Quote Originally Posted by freakymath View Post
    2) y'' – 2y' + y = x- ex
    This one can't be too tricky, have you tried to solve the non homogenous version of this D.E?

    It may be worth reading this thread http://www.mathhelpforum.com/math-he...ial-38182.html
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  3. #3
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    Quote Originally Posted by freakymath View Post
    I have 3 questions and NO TIME!
    1) xy'' + xy' - y = xlnx maybe about euler
    For this one, you can use the power series substitution y = \sum_{n=0}^\infty a_n x^n
    And set this equal to the Taylor expansion of x\ln(x) to find the coefficients a_n
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  4. #4
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    I solved it. What about the first one: I solved it too but the other side of the equation is =0 NOT =xlnx. What can i do about that problem?
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  5. #5
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    Sorry I'm not a student of math department. So can you explain ur ideas deeply with some expressions? Pls...
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  6. #6
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    So, the first question is x^2 y'' + xy' + y = 0 ?

    Well, then let
     y = \sum_{n=0}^\infty a_nx^n
    y' = \sum_{n=0}^\infty a_nnx^{n-1}
    y'' = \sum_{n=0}^\infty a_nn(n-1)x^{n-2}

    x^2 y'' + xy' + y = \sum_{n=0}^\infty((n+1)(n+2)a_{n+2} + (n+1)a_{n+1} + a_n)x^n = 0

    This means:
    (n+1)(n+2)a_{n+2} + (n+1)a_{n+1} + a_n = 0
    Now you have the recurrence
    a_{n+2} = -\frac{a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}

    You need the initial conditions to find a_0 and a_1.
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  7. #7
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    No it's xy'' + xy' - y = xlnx
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  8. #8
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    snowtea; i used euler. But at the end there is xlnx. So what will i do at the end???
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  9. #9
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    If it is =0, the result is y(x)=c1(first c)x+c2(second c)/x
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  10. #10
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    I'm not sure what you mean by Euler. Do you mean Euler's method to numerically estimate the function? If not, I do not know enough to help you on problem 1, sorry.
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  11. #11
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    euler is what you had written y,y',y'' ...
    but the main point is xlnx here. im a rookie, pls forgive me.
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  12. #12
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    Ok, I see. Did not know about this type of differential equation was also Euler's.
    The way an Euler differential equation is transformed into a linear ODE is by substituting x = e^t or t = \ln(x).

    So substitue this for x on the right hand side.

    x\ln(x) = te^t

    The new differential equation is (where all the derivatives are now w.r.t. t):

    (y'' - y') + y' - y = te^t
    y'' - y = te^t

    Solve this differential equation to get something of the form:
    y = Ae^{t} + Be^{-t} + Cte^t

    Now substitute in t = \ln(x) to get the final solution.

    y = Ax + \frac{B}{x} + Cx\ln(x)
    Last edited by snowtea; December 17th 2010 at 06:13 PM.
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  13. #13
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    ImageShack® - Online Photo and Video Hosting
    See what i did under the line for =0. I think it's true.
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  14. #14
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    Yes, for x^2y'' + xy' + y =0, you are correct.
    For x^2y'' + xy' + y = xlnx, just substitute x=e^t on the left hand side also (see what I posted above).
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  15. #15
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    Lemme try it ...
    Last edited by mr fantastic; December 17th 2010 at 09:25 PM.
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