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Math Help - find the values of m

  1. #1
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    find the values of m

    ok i wanted to know if someone could check my work and let me know if they got the same thing.

    im supposed to find the values of m so that the function y = x^m is a solution to,

    x^2y'' - 3xy' - y = 0.

    i got m = 0. i just took the first and 2nd derivative of the function and plugged it into the equation and solved.

    thanks in advance.
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  2. #2
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    You should get : m=2 \pm \sqrt{5}.

    so that the two solutions are : y_1=x^{2+\sqrt{5}} & y_2=x^{2-\sqrt{5}}.

    Hence, the general solution is y=c_1x^{2+\sqrt{5}}+c_2x^{2-\sqrt{5}}.
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  3. #3
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    yeah i was or am a little tired, ive been trying to get ready for this exam all night..lol thanks for your reply.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Generalization :

    In case we don't know there are solutions of the form y=x^m and considering that we have an Euler equation, the subtitution x=e^u with u independent variable, transforms the given equation into a linear homogeneous differential equation with constant coefficients.



    Fernando Revilla
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