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Math Help - D.e inhomogneous non constant coeff

  1. #1
    Member i_zz_y_ill's Avatar
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    D.e inhomogneous non constant coeff

    The question is find the solution of:

    x^2\frac{dy}{dx}-x\frac{dy}{dx}-8y=1+x

    clearly the coefficients in what i would write down to be the auxilliary equation are not constants. Ive been told by my solution to use y= x^m but i dont see how this gives(in the solutions) m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0
    giving m=4 or m=-2. and the corrosponding complementary function. Does anyone have any ideas or could show me how the method actually cancels out the non constant coefficeints? thanks. oh i think you are supposed to use liebnitz?? not sure
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  2. #2
    Senior Member Spec's Avatar
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    Cauchy-Euler equation - Wikipedia, the free encyclopedia

    One way to solve it is to set t=\ln x,\ x > 0 (if x < 0 use t=\ln -x instead)

    Let y(x)=z(t(x))

    Then \frac{dy}{dx}=\frac{dz}{dt}\frac{dt}{dx}=\frac{1}{  x}\frac{dz}{dt} and xy'(x)=z'(t)

    \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\fr  ac{dz}{dt} \right)=-\frac{1}{x^2}\frac{dz}{dt}+\frac{1}{x}\frac{d}{dx}  \left( \frac{dz}{dt}\right)=-\frac{1}{x^2}\frac{dz}{dt}+\frac{1}{x^2}\frac{d^2z  }{dt^2}

    We have that: x^2y''(x)=z''(t)-z'(t)

    So instead you need to solve z''(t)-2z'(t)-8z(t)=1+e^t
    Last edited by Spec; May 18th 2009 at 07:06 AM. Reason: Wikipedia URL
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  3. #3
    MHF Contributor chisigma's Avatar
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    The complete equation is...

    x^{2}\cdot y^{''} - x \cdot y^{'} - 8 \cdot y = 1 + x (1)

    Both x^{-2} and x^{4} are solution of the incomplete equation and they are also linearly independent, so that the general integral of the incomplete equation is ...

    y(x)= c_{1}\cdot x^{-2} + c_{2}\cdot x^{4} (2)

    With little patience you find that -\frac {1}{8} - \frac{x}{9} is solution of (1) , so that the general integral of (1) is...

    y(x)= c_{1}\cdot x^{-2} + c_{2}\cdot x^{4} -\frac {1}{8} - \frac{x}{9} (3)

    Kind regards

    \chi \sigma
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  4. #4
    Member i_zz_y_ill's Avatar
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    in the solution they used y=x^m which is entirely different from m=lnx which is the equivalent of what your saying..im a little confused as to how you did \frac{d}{dx}(\frac{1}{x}\frac{dz}{dt}) aswell? ''.
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  5. #5
    Member i_zz_y_ill's Avatar
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    i think the easist way would be if someone showed me how they did it with y=x^m
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  6. #6
    MHF Contributor chisigma's Avatar
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    You have to consider as first step the searching ot solution of the incomplete equation...

     x^{2}\cdot y^{''} - x \cdot y^{'} - 8 \cdot y = 0 (1)


    If you suppose that x^{m} is solution of (1) , you find by simple substitution in (1) that that is true for m=-2 and m=4. But they are linearly independent so that the general integral of (1) is...

    y(x) = c_{1}\cdot x^{-2} + c_{2}\cdot x^{4} (2)

    The first step is terminated. The second step consists in searching a particular solution of the complete equation... not a very difficult task...

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    \chi \sigma
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  7. #7
    Member i_zz_y_ill's Avatar
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    I dont underrstand could you just show me how to do it using y=x^m
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  8. #8
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    Quote Originally Posted by i_zz_y_ill View Post
    The question is find the solution of:

    x^2\frac{dy}{dx}-x\frac{dy}{dx}-8y=1+x

    clearly the coefficients in what i would write down to be the auxilliary equation are not constants. Ive been told by my solution to use y= x^m but i dont see how this gives(in the solutions) m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0
    giving m=4 or m=-2. and the corrosponding complementary function. Does anyone have any ideas or could show me how the method actually cancels out the non constant coefficeints? thanks. oh i think you are supposed to use liebnitz?? not sure
    If you use the anzats y=x^m and take two derivatives we get

    y'=mx^{m-1} and

    y''=m(m-1)x^{m-2} Now if you plug this into the homogenious ODE you get

    x^2(m(m-1)x^{m-2})-x(m(x^{m-1}))-8x^m=0

    lets reduce to get

    m(m-1)x^m-mx^{m}-8x^m=0 \iff [m(m-1)-m-8]x^m=0

    Since x^m \ne 0 \implies m(m-1)-m-8=0

    Solving we get m^2-2m-8=0 \iff (m-4)(m+2)=0 so

    m=4 \mbox{ or } m=-2 Plug these back into your anzats and you have your complimentry solution

    y_c=c_1x^{4}+c_2x^{-2}
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