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Thread: 2nd order ODE

  1. #1
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    2nd order ODE

    Hi ,
    I am stuck with the following problem (to turn tomorrow):

    Find a second order linear homogeneous equation having the pair as a fundamental set of solutions:
    y1(x)=x , y2(x)=x*ln(x).


    My problem here is that I dont have the exponential form for the proposed solutions.

    Thank you for your help

    B.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy
    Hi ,
    I am stuck with the following problem (to turn tomorrow):

    Find a second order linear homogeneous equation having the pair as a fundamental set of solutions:
    y1(x)=x , y2(x)=x*ln(x).


    My problem here is that I dont have the exponential form for the proposed solutions.

    Thank you for your help

    B.
    A second order linear homogeneous ODE has the form:

    $\displaystyle
    \ddot y + f(x) \dot y+g(x) y=0
    $

    Now

    $\displaystyle
    y(x)=x
    $

    is a solution so differentiating twice and substituting the derivatives of $\displaystyle y$
    into the ODE gives:

    $\displaystyle
    f(x)+xg(x)=0
    $

    or:

    $\displaystyle
    g(x)=-\frac{f(x)}{x}\ \ \ ...(1)
    $

    Similarly

    $\displaystyle
    y(x)=x \ln(x)
    $

    is also a solution so:

    $\displaystyle
    \frac{1}{x}+(\ln (x)+1)f(x)+x \ln (x)g(x)=0
    $

    Now substituting $\displaystyle g(x)$ from $\displaystyle (1)$ into this last
    equation gives:

    $\displaystyle
    \frac{1}{x}+(\ln (x)+1)f(x)- \ln (x)f(x)=0
    $

    So rearranging we find:

    $\displaystyle
    f(x)=-\frac{1}{x}
    $

    and:

    $\displaystyle
    g(x)=\frac{1}{x^2}
    $

    RonL
    Last edited by CaptainBlack; Feb 15th 2006 at 07:02 AM.
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  3. #3
    Member
    Joined
    Nov 2005
    Posts
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    Quote Originally Posted by CaptainBlack
    A second order linear homogeneous ODE has the form:

    $\displaystyle
    \ddot y + f(x) \dot y+g(x) y=0
    $

    Now

    $\displaystyle
    y(x)=x
    $

    is a solution so differentiating twice and substituting the derivatives of $\displaystyle y$
    into the ODE gives:

    $\displaystyle
    f(x)+xg(x)=0
    $

    or:

    $\displaystyle
    g(x)=-\frac{f(x)}{x}\ \ \ ...(1)
    $

    Similarly

    $\displaystyle
    y(x)=x \ln(x)
    $

    is also a solution so:

    $\displaystyle
    \frac{1}{x}+(\ln (x)+1)f(x)+x \ln (x)g(x)=0
    $

    Now substituting $\displaystyle g(x)$ from $\displaystyle (1)$ into this last
    equation gives:

    $\displaystyle
    \frac{1}{x}+(\ln (x)+1)f(x)- \ln (x)f(x)=0
    $

    So rearranging we find:

    $\displaystyle
    f(x)=-\frac{1}{x}
    $

    and:

    $\displaystyle
    g(x)=\frac{1}{x^2}
    $

    RonL
    Thank you very much Captain,
    Because we use a different form in class, the exponential form for the solution (y=e(rx)), I was a little bit confused by this problem . Now I get it!

    B
    Last edited by CaptainBlack; Feb 15th 2006 at 07:03 AM.
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