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Math Help - 2nd order DE (non linear I think but homogeneous)

  1. #1
    MHF Contributor arbolis's Avatar
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    2nd order DE (non linear I think but homogeneous)

    Find the general solution of x^2y''-5xy'+9y=0. Hint: Once known a solution y_1(x), another solution is given by y(x)=v(x)y_1(x), with v(x) to determinate.
    Attempt: I just don't know what method to use. I don't think variation of parameters work here since it's non linear.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Its a Cauchy-Euler equation.

    So suppose that the equation ax^2y^{\prime\prime}+bxy^{\prime}+cy=0 has a solution of the form y=x^r. Plugging this back into the ODE yields the characteristic equation ar^2+(b-a)r+c=0.

    In this particular problem, x^2y^{\prime\prime}-5xy^{\prime}+9y=0 has the characteristic equation r^2-6r+9=0, which yields repeated roots.

    So once you have y_1(x), you suppose y_2(x)=u(x)y_1(x). It turns out that once you write the ODE in the form y^{\prime\prime}+P(x)y^{\prime}+Q(x)y=0 and substitute y_2(x)=u(x)y_1(x) into it, we see that u(x)=\displaystyle\int\left[\frac{e^{-\int P(x)dx}}{[y_1(x)]^2}\right]dx (this is the reduction of order technique as seen in my tutorial here.)

    Then at this point, we see that y_2(x)=\left(\displaystyle\int\left[\frac{e^{-\int P(x)dx}}{[y_1(x)]^2}\right]dx\right)y_1(x) is the second solution.

    Can you continue with the problem?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks for the help, I will try. However isn't the characteristic equation r^2-5r+9=0 instead of r^2-6r+9=0? So that the roots aren't repeated. I should solve both problems anyway.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Let a=1, b=-5 and c=9. In this particular case, it has to be r^2+(-5-1)r+9=r^2-6r+9=0. Observe that ax^2y^{\prime\prime}+bxy^{\prime}+cy=0 and ay^{\prime\prime}+by^{\prime}+cy=0 have different characteristic equations since they assume different solutions (the former assumes solutions of the form y=x^r and the latter assumes solutions of the form y=e^{rx})!

    The former type of DE has the characteristic equation ar^2+(b-a)r+c=0, where as the latter type has the equation ar^2+br+c=0.
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  5. #5
    MHF Contributor arbolis's Avatar
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    I should have read better your first reply.
    Ok, I get y_1 (x)=x^3.
    With P(x)=-\frac{5}{x}, I reach that u(x)=k \log |x| +C.
    Thus the general solution would be of the form y(x)=x^3 (c_1 \log |x| + c_2)+x^3=x^3(c_1 \log |x| + c_3).
    I hope it's right. And sorry, I didn't recognize it was a Cauchy-Euler equation.
    Last edited by arbolis; August 1st 2010 at 07:30 PM. Reason: I forgot to add the first solution
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I should have read better your first reply.
    Ok, I get y_1 (x)=x^3.
    With P(x)=-\frac{5}{x}, I reach that u(x)=k \log |x| +C.
    Thus the general solution would be of the form y(x)=x^3 (c_1 \log |x| + c_2)+x^3=x^3(c_1 \log |x| + c_3).
    I hope it's right. And sorry, I didn't recognize it was a Cauchy-Euler equation.
    No need to be sorry! Its just something that is useful to know down the road. Now, when we go ahead and find u(x), we don't care much for the constants (they should drop out in the end anyways); so what where really after is that u(x)=\log x (or u(x)=\ln x, depending on what notation is used where you're at). Thus, we see that the general solution is y=c_1x^3+c_2x^3\log x, which is what you have for your solution (but written in a slightly different way).

    Hope that clarifies things!
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