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Math Help - 2nd order diff equation!

  1. #1
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    2nd order diff equation!

    Hi, I have this differential equation to solve its broken down in to 3 parts but i`m having trouble doing the 2nd part any ideas would be great!

    Consider the intial value problem

    (y`` x^2) - (4xy`) + 6y = g(x) x>0 y(0)=y`(0)=0

    part a) Write this de in self adjoint form which i have done

    (1/x^4 y`)` + 6y/x^6 = g(x)x^-6

    part b) Find the influence function for this

    i know the formula for this but i have to find y1 and y2 but i have no idea on how to do this can anyone help me?

    thanks
    Last edited by FunkyBudduh; October 27th 2010 at 04:03 PM. Reason: wrong thread =(
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  2. #2
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    Please try this

    <br />
\displaystyle {y=x^r<br />
}<br />
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  3. #3
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    Quote Originally Posted by FunkyBudduh View Post
    (y`` x^2) - (4xy`) + 6y = g(x) x>0 y(0)=y`(0)=0
    put t=\ln x then \displaystyle\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}, then \displaystyle\frac{{{d}^{2}}y}{d{{t}^{2}}}-5\frac{dy}{dt}+6y(t)=g({{e}^{t}}).

    homogeneous solution is y_h(t)=c_1e^{3t}+c_2e^{2t} now find the particular solution.
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  4. #4
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    Hi Krizalid,

    Why is it -5dy/dx?

    thanks
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  5. #5
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    Krizalid's Avatar
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    it's dt not dx.
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  6. #6
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    Hi Krizalid,

    Why are we using the function t=ln(x) instead of y=x^r seeming as the LHS of the equation is a polynomial and not a exponential. When i have used y= x^r i have got y=x^2 and y=x^3.

    thanks
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  7. #7
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    it was a matter of choice, both ways work.
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  8. #8
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    Hi Krizalid,

    part 3 of the question is that g(x) =  (x^4)(e^x)

    i have used y=x^2 and y=x^3
    to obtain a really nasty general solution, with exponentials and powers of x`s

    it looks like

    e^x(-x^7 + 8x^6 - 48x^5.....

    pretty sure this is wrong though, any idea on how to solve this i been tryin to do this for days now ...

    thanks

    much appreciated
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  9. #9
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    I HAVE SOLVED THIS!!!!!!!!!!!!!!! Thx For your help all
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