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Homework Help: Higher Order Non-Homogenous ODE

  1. Apr 19, 2006 #1
    Higher Order Homogenous ODE (Euler-Cauchy)

    *sigh* I am (yet again) stuck on a problem.. I would greatly appreciate any help!

    [tex] x^3y''' - 3x^2y'' + (6-x^2)xy' - (6-x^2)y = 0 [/tex]

    [tex] \inline y_1 = x [/tex] is a solution to the equation above

    y'(0) = 3
    y''(0) = 9
    y'''(0) = 18

    I'm not quite sure what it means by "[tex] \inline y_1 = x [/tex] is a solution to the equation above" so I decided to ignore it. I probably shouldn't have because it is probably something I need to know and it is probably the reason why I haven't been able to solve this problem.

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    What I have done so far

    Substituting [tex] \inline m^n [/tex] for y and dividing by [tex] \inline x^m [/tex] I got:

    [tex] m(m-1)(m-2) - 3m(m-1) + (6-x^2)m - (6-x^2) = 0 [/tex]

    [tex] m(m-1)(m-2) - 3m(m-1) + (6-x^2)(m-1) = 0 [/tex]

    [tex] m(m-2) - 3m + (6-x^2) = 0 [/tex]

    [tex] m^2 - 5m + (6-x^2) = 0 [/tex]

    It is at this point that I am confused. It looks like a quadratic equation but what do I do with the (6-x^2)?!?


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    Question 1:

    What DO I do with (6-x^2) ?!


    Someone told me that he got the roots 1, 2, 3... which I then tried, but got the answer wrong:

    [tex] y = c_1x + c_2x^2 + c_3x^3 [/tex]

    [tex] y' = c + 2c_2x + 3c_3x^2 [/tex]

    [tex] y'' = 2c_2 + 6c_3x [/tex]

    [tex] y''' = 6c_3 [/tex]

    And putting in the initial values I get:

    [tex] c_1 = 3 [/tex]

    [tex] c_2 = 9/2 [/tex]

    [tex] c_3 = 3 [/tex]

    Which doesn't seem right.. (can there be two different constants that have the same value?) But I put in the found constants into the original equation to get:

    [tex] y = 3x + \frac{9}{2}x^2 + 3x^3 [/tex]

    Which is.. wrong.


    -----------
    Question 2:

    Are the roots not 1, 2, 3 or did I do something wrong?
     
    Last edited: Apr 20, 2006
  2. jcsd
  3. Apr 20, 2006 #2

    benorin

    User Avatar
    Homework Helper

    Use the substitution [tex]y=xu[/tex] where the x is from the known solution and u is a function of x [i.e. u=u(x)], we also have

    [tex]y=xu[/tex]
    [tex]y^{\prime}=xu^{\prime}+u[/tex]
    [tex]y^{\prime\prime}=xu^{\prime\prime}+2u^{\prime}[/tex]
    [tex]y^{\prime\prime\prime}=xu^{\prime\prime\prime}+3u^{\prime\prime}[/tex]

    substitute these into the given 3rd order DE to get (after a good deal of cancelation)

    [tex]x^3u^{\prime\prime\prime}-x^4u^{\prime}=0[/tex]
     
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