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Math Help - Second Order linear homog ODE

  1. #1
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    Second Order linear homog ODE

    Find two linearly independent solutions of

    x2y''+x y'' + y = 0

    Also find the solutions which satisfy

    y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,


    So subbing in y(x)= xr

    You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,

    y1 satisfies the solutions provided by y2 does not,

    The y2 solutions look like they're for ln(x)

    Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

    Thanks

    Linda
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  2. #2
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    Re: Second Order linear homog ODE

    Quote Originally Posted by Linda200 View Post
    Find two linearly independent solutions of

    x2y''+x y'' + y = 0

    Also find the solutions which satisfy

    y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,


    So subbing in y(x)= xr
    You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,
    No, you don't. If y= xr then y'= rxr-1 and y''= r(r-1)xr-2. Putting that into your equation gives
    r(r-1)xrr+ xr= xr(r2+ 1)= 0.

    In order that that be true for all x, you must have r2+ 1= 0 which does NOT have 1 and -1 as roots.

    A different way of solving an equation like this is to make the change of variable x= et which changes the d.e. into a "constant coefficients" equation that has the same characteristic equation.

    y1 satisfies the solutions provided by y2 does not,
    No, neither satisfies the equation. If y= x then y'= 1, y''= 0 so your equation becomes x+ x= 2x, not 0.

    The y2 solutions look like they're for ln(x)

    Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

    Thanks

    Linda[/QUOTE]
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  3. #3
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    Re: Second Order linear homog ODE

    x2y''+x y'' - y = 0

    I miswrote the equation,

    I am very sorry,

    I had a look at that equation for practice though and I got solutions that worked,

    since its xi, x-i i used eulers formula to turn it into exp[i ln(x)] which is cos + isin
    so the solutions are y=cos(ln(x)) and y2=sin(ln(x)) which is consistent with the initial conditions

    but with
    x2y''+x y'' - y = 0
    i can't see a way to get x and x-1into a form which satisfies the initial conditions
    Last edited by Linda200; March 31st 2012 at 05:24 PM.
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  4. #4
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    Re: Second Order linear homog ODE

    Quote Originally Posted by Linda200 View Post
    Find two linearly independent solutions of

    x2y''+x y'' + y = 0

    Also find the solutions which satisfy

    y1(1) = 1, y'1(1)=0, y2(1) = 0, y'2(1)=1,


    So subbing in y(x)= xr

    You get solutions y1(x) = x and y2(x)=x-1 which are linearly independent,

    y1 satisfies the solutions provided by y2 does not,

    The y2 solutions look like they're for ln(x)

    Does anyone know how I would get ln(x) as a solutions from x-1? (I know ln(x) is just the integrate of x-1 but I don't know why you would integrate)

    Thanks

    Linda
    Is your original DE \displaystyle \begin{align*} x^2y'' + x\,y'' + y = 0 \end{align*} or \displaystyle \begin{align*} x^2y'' + x\,y' + y = 0 \end{align*}?
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  5. #5
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    Re: Second Order linear homog ODE

    The two odes your asked about are the same,

    the original was x2y''+x y'' - y = 0

    I solved it earlier today so it doesn't,

    My mistyped equation was good for practice anyway he he

    Thanks for your help guys!
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