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Thread: 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:

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

    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=$\displaystyle x^m$ but i dont see how this gives(in the solutions) $\displaystyle m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0$
    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 $\displaystyle t=\ln x,\ x > 0$ (if $\displaystyle x < 0$ use $\displaystyle t=\ln -x$ instead)

    Let $\displaystyle y(x)=z(t(x))$

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

    $\displaystyle \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: $\displaystyle x^2y''(x)=z''(t)-z'(t)$

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    Member i_zz_y_ill's Avatar
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    in the solution they used $\displaystyle y=x^m$ which is entirely different from m=lnx which is the equivalent of what your saying..im a little confused as to how you did $\displaystyle \frac{d}{dx}(\frac{1}{x}\frac{dz}{dt})$ 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 $\displaystyle y=x^m$
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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...

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


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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle y=x^m$
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  8. #8
    Behold, the power of SARDINES!
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    Quote Originally Posted by i_zz_y_ill View Post
    The question is find the solution of:

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

    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=$\displaystyle x^m$ but i dont see how this gives(in the solutions) $\displaystyle m(m-1)-m-8=x^2-2m-8=(x-4)(x+2)0$
    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 $\displaystyle y=x^m$ and take two derivatives we get

    $\displaystyle y'=mx^{m-1}$ and

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

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

    lets reduce to get

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

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

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

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

    $\displaystyle y_c=c_1x^{4}+c_2x^{-2}$
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